Time to cover first 40 km is 40/20 = 2 hours .
Total time to cover total distance of 80 km is 80 /30 =
8/3 hours .
Time to cover second 40 km is 8/3 - 2 = 2/3 hours .
Second speed is 40 / ( 2/3) = 20 *3 = 60 km/h
Let the speed for the second 40 km be x kmph
AVERAGE SPEED = TOTAL DISTANCE / TOTAL TIME
TOTAL DISTANCE = 40+40 = 80 km
Total Time = 40/20 + 40/x = 2 +40/x hours
30 = 80 / (2+ 40/x)
60 + 1200/x = 80
1200/x = 20
x= 1200/20= 60 kmph ANSWER
CHECK
40/20= 2 hours
40/60 = 2/3 hours
Total time = 2 2/3 hours
80 / 2 2/3 = 80 x 3/8 = 30 kmph
Where do I start?
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Where do I start?
I’m a huge financial nerd, and have spent an embarrassing amount of time talking to people about their money habits.
Here are the biggest mistakes people are making and how to fix them:
Not having a separate high interest savings account
Having a separate account allows you to see the results of all your hard work and keep your money separate so you're less tempted to spend it.
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To find the speed of the motorcycle for the next 40 km journey so that the average speed for the entire 80 km journey is 30 kph, we can use the formula for average speed:
[math]\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}[/math]
Step 1: Calculate Total Distance
The total distance of the journey is:
[math]\text{Total Distance} = 40 \text{ km} + 40 \text{ km} = 80 \text{ km}[/math]
Step 2: Calculate Required Total Time
To achieve an average speed of 30 kph over 80 km, we can find the required total time:
[math]\text{Total Time} = \frac{\text{Total Distance}}{\text{Average Speed}} = \frac{80 \text{ km}}{30 [/math]
To find the speed of the motorcycle for the next 40 km journey so that the average speed for the entire 80 km journey is 30 kph, we can use the formula for average speed:
[math]\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}[/math]
Step 1: Calculate Total Distance
The total distance of the journey is:
[math]\text{Total Distance} = 40 \text{ km} + 40 \text{ km} = 80 \text{ km}[/math]
Step 2: Calculate Required Total Time
To achieve an average speed of 30 kph over 80 km, we can find the required total time:
[math]\text{Total Time} = \frac{\text{Total Distance}}{\text{Average Speed}} = \frac{80 \text{ km}}{30 \text{ kph}} = \frac{8}{3} \text{ hours} \approx 2.67 \text{ hours}[/math]
Step 3: Calculate Time Taken for the First 40 km
The time taken to cover the first 40 km at 20 kph is:
[math]\text{Time}_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{40 \text{ km}}{20 \text{ kph}} = 2 \text{ hours}[/math]
Step 4: Calculate Remaining Time for the Next 40 km
Now, we find the remaining time for the second part of the journey:
[math]\text{Remaining Time} = \text{Total Time} - \text{Time}_1 = \frac{8}{3} \text{ hours} - 2 \text{ hours} = \frac{8}{3} - \frac{6}{3} = \frac{2}{3} \text{ hours}[/math]
Step 5: Calculate Required Speed for the Next 40 km
To find the speed needed to cover the next 40 km in [math]\frac{2}{3}[/math] hours, we use the formula:
[math]\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{40 \text{ km}}{\frac{2}{3} \text{ hours}} = 40 \text{ km} \times \frac{3}{2} = 60 \text{ kph}[/math]
Conclusion
The speed of the motorcycle for the next 40 km journey should be 60 kph to achieve an average speed of 30 kph for the entire 80 km journey.
Let Vavg be average speed and v2 be x km/hr
Here simply apply the formula for average speed having speed v1,v2= 2v1v2/(v1+v2)
Here v1=20 km/hr
v2=??
Vavg= 30 km/hr
Now applying formula
Vavg= 2v1v2/(v1+v2)
2×20×x/(20+x)=30
40x =600+30x
10x=600
x= 60 km/hr
Hence the speed of motorcycle for next 40 km
= 60 km/hr
Speed of the motorcycle for the first 40 kms is 20 km/h
And let speed of the motorcycle for the next 40 km of journey be 'x' km/h
Now given average speed of total journey is 30 km/h
So V@avg =total distance travelled ÷ total time taken
So 30 = 80/(40/20 + 40/x)
Solving we get x = 60 km/h
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Mos
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- Total distance travelled= (40+40)= 80- km
- Average speed= 30- Kms/ h.
- (Time) Duration of travel = 80/30 =8/3hours
- So the next 40- Kms should be travelled in (8/3- 2)= ⅔- hours.
- So the speed of the motorcycle for the next 40- Kms = 40/(⅔)
- = 60- Km/ h
- According to given problem,
- (i) A motorcycle covers 50 km with a speed of 25 km/hr.
- (ii) Let S denotes the speed of the motorcycle for the next 40 km journey so that the average speed of the whole journey will be 20 km/hr.
- We know that,
- Average Speed = (Total travel-distance)/(Total travel-time) …… (1a)
- Therefore from (1a), (i) & (ii) we get following relation,
- 20 = (50 + 40)/(50/25 + 40/S) = 90/(2 + 40/S)
- or 2*(2 + 40/S) = 9 or 4 + 80/S = 9 or 80/S = 5 or S = 16 (km/hr) [Ans]
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Let, the speed of the motorcycle for the next 40 km is x kmph
T1 =48/24=2 hours
T2= 40/x hours
And, S1+S2 =48+40 =88 km
(S1+S2) /(T1+T2) = 20
Or, 88/(2+40/x)=20
Or, 88x=20(2x+40)
Or, 88x =40x +800
Or, 48x =800
Or, x=800/48=100/6=50/3=16.66666 (approximately)
So, the speed of the motorcycle for the next 40 km is =16.667 kmph (ans)
Total distance 40+40=80kms
av speed = 30kms/hr
total time =80/30=2 2/3 hrs
First 40 kms travelled at 20 km/hr in time=40/20=2 hrs
So second 40 kms travelled in 2 2/3–2=2/3 hours
Speed in travelling 2nd 40 kms=40/(2/3)
=60 km/hr
ans-16 km/h
average speed=distance covered/time taken.
total distance=50+40=90km
average speed=20k/h
total time taken=distance/speed=90/20=4.5 hr
in first case time taken= 50/25=2 hr
time taken for second half of journey=4.5–2=2.5hr
so speed of motorcycle in second half of journey=40/2.5=16k/h
Total journey=(40+40)km=80km.
Average speed 30km/h
Total time 80/30h=8/3h
Time to cover first 40km=40/20h=2h
Time left=(8/3–2)h=2/3h.
Speed for the next 40km=40÷2/3km/h=60km/h
Ans:- The speed of the next 40km should be 60km/h.
It can be solved by using simple formula 2v1v2/v1+V2
Check option if you are solving mcq type.
For method…..
Total time=80/30=8/3hr
8/3=40/20+t
t=2/3 HR
Speed= 40/2*3=60km/hr
AVERAGE SPEED = TOTAL DISTANCE / TOTAL TIME
TOTAL DISTANCE = 50+40 = 90 km
AVERAGE SPEED = 20 km /h
TOTAL TIME = TOTAL DISTANCE / AVERAGE SPEED = 90/20 = 9/2 = 4.5 HOURS
50 /25 = 2 HOURS
SO 40 KM WAS COVERED IN 4.5–2 = 2.5 HOURS
SPEED = 40/2.5 = 16 km/h ANSWER
Time for 80km journey=80/30=8/3 hrs. Time already passed =40/20=2 hrs . So time left is =8/3–2=2/3 hrs. So speed during next 40 km =40/(2/3)=60 km/hr.
[math]\text{Average speed is, }v_a=\dfrac{\text{Total distance travelled} }{\text{Total time taken}}[/math]
[math] \text{Data: Total Distance }s=25 km +20 km =45 km, \text{ average speed, } v_a= 10 km/h[/math]
[math] \text{Accordingly, } v_1=50 km/h, v_2=20 km/t h[/math]
[math] v_a=\dfrac{s}{T}[/math]
[math] \text{We know that }s=55km, T=t_1+t_2[/math]
[math] t_1=\dfrac{25 km}{50 km/h}, t_2=\dfrac {20 km}{20 km/th}[/math]
[math] t_1=0.5h, t_2=t h[/math]
[math] 1...
Total distance to cover=25+20=45 km and time available is 45/10=4.5 hrs =4 hr 30 minutes. Distance left =20 km and tome left =4.5-(25/50)=4 hrs. So avg speed for remaining journey is 20/4=5 km/hr.
- Statement of the given problem,
- A motorcycle covers 25 km with a speed of 50 kph. What is the speed of the motorcycle for the next 20 km journey so that the average speed of a whole journey will be 10 kph?
- Let S denotes the required speed (in kph) of the motorcycle for the next 20 km journey.
- We know that
- (Total travel-distance)/(Total travel-time) = Average speed
- Hence from above data we get following relation,
- (25 + 20)/(25/50 + 20/S) = 10
- or 45 = 5 + 200/S or 200/S = 40 or S = 200/40 = 5 (kph) [Ans]
Wow, sounds like a big motorcycle to be able to cover that much land! I’m assuming you’re referring to square kilometers, so 2.4 km^2 in five minutes is really fast. Let’s do the nitty-gritty:
Let S = number of motorcycle parts
Let B = square meter per square kilometer (m^2/km^2)
Let u = average surface area of each part (m^2)
Let r = the radius of the wheel (m)
Let p = the average shininess of the parts of the motorcycle (Coloumbs/sec)
Let a = area of the motorcycle before the devastating accident (m^2)
Let m = mass
Let y = why am I doing this to myself
Let o = oh yeah I’ve got nothing better to do an
Wow, sounds like a big motorcycle to be able to cover that much land! I’m assuming you’re referring to square kilometers, so 2.4 km^2 in five minutes is really fast. Let’s do the nitty-gritty:
Let S = number of motorcycle parts
Let B = square meter per square kilometer (m^2/km^2)
Let u = average surface area of each part (m^2)
Let r = the radius of the wheel (m)
Let p = the average shininess of the parts of the motorcycle (Coloumbs/sec)
Let a = area of the motorcycle before the devastating accident (m^2)
Let m = mass
Let y = why am I doing this to myself
Let o = oh yeah I’ve got nothing better to do and I’m alone
Using Pascal’s first law of Pascal’s triangle:
O^2r^2sy/ = BrMa/S
- or -
Supra = MayroBros
Since the motorcycle is moving”along a straight Eastward road@ a constant speed”, its acceleration is zero. And so, moving with this uniform velocity,it will cover a distance of S=V t= 12*300= 3600 m in this five minute.
AFTER SOLVING YOUR PROBLEM,I HAVE POSTED A PAIR OF VIDEOS WITH THE FIRST ONE PERTAINING TO THE PROBLEM STATEMENT WHILE THE SECOND ONE TO THE SOLUTION. THESE ARE JUST TWO, OUT OF MANY SUCH VIDEOS AVAILABLE ON MY YOUTUBE CHANNEL CALLED THE”RCM Science Channel”. ON THIS CHANNEL MANY INTERESTING PROBLEMS ON PHYSICS ALONG WITH
Since the motorcycle is moving”along a straight Eastward road@ a constant speed”, its acceleration is zero. And so, moving with this uniform velocity,it will cover a distance of S=V t= 12*300= 3600 m in this five minute.
AFTER SOLVING YOUR PROBLEM,I HAVE POSTED A PAIR OF VIDEOS WITH THE FIRST ONE PERTAINING TO THE PROBLEM STATEMENT WHILE THE SECOND ONE TO THE SOLUTION. THESE ARE JUST TWO, OUT OF MANY SUCH VIDEOS AVAILABLE ON MY YOUTUBE CHANNEL CALLED THE”RCM Science Channel”. ON THIS CHANNEL MANY INTERESTING PROBLEMS ON PHYSICS ALONG WITH THEIR SOLUTIONS ARE AVAILABLE. HERE EACH PROBLEM IS FOLLOWED BY ITS SOLUTION. THESE ARE VERY HELPFUL FOR NEGOTIATING YOUR HOMEWORK AS WELL AS FOR LEARNING THE SUBJECT OF PHYSICS. YOU SHOULD WATCH AND SUBSCRIBE AND SHOULD ALSO FORWARD THOSE VIDEOS TO OTHERS AS WELL.
Answer to the question as asked: There are infinitely many answers, arising from infinitely many possibilities for multiple unknown factors. To give just two examples: what was the motorcycle’s initial velocity, and exactly how hard was it accelerating and/or decelerating at every instant over the first 5 seconds?
Some additional conditions: Here’s one possible set of additional conditions that will yield a unique answer. Assume that the motorcycle moves in a straight line — with constant acceleration — for the first 5 seconds. I can think of both a conceptual approach (not requiring calculus)
Answer to the question as asked: There are infinitely many answers, arising from infinitely many possibilities for multiple unknown factors. To give just two examples: what was the motorcycle’s initial velocity, and exactly how hard was it accelerating and/or decelerating at every instant over the first 5 seconds?
Some additional conditions: Here’s one possible set of additional conditions that will yield a unique answer. Assume that the motorcycle moves in a straight line — with constant acceleration — for the first 5 seconds. I can think of both a conceptual approach (not requiring calculus) and a calculus approach to this modified problem.
Conceptual approach:
Velocity is change in position over time. If the motorcycle is constantly accelerating, that means its true or instantaneous velocity is continuously increasing. But we can easily calculate its average velocity over the time intervals mentioned in the question, using the definition of velocity. For the first two seconds, the motorcycle’s average velocity (i.e change in position over time) was [math]20 m / 2 sec = 10 m/s[/math]. For the next 2 seconds, it was [math]40 m / 2 sec = 20 m/s[/math].
Acceleration is change in velocity over time, so the motorcycle’s acceleration was [math](20 m/s - 10 m/s) / (3 sec - 1 sec) = 5 m/s/s[/math]. Why the 3 and 1? Keep reading.
What does an acceleration of [math]5 m/s/s[/math] mean? It means that every second that goes by, the motorcycle’s instantaneous velocity hits [math]5 m/s[/math] higher. So 1 second after its velocity hit [math]10 m/s[/math], it hit [math]15 m/s[/math]. And 1 second before it hit [math]10 m/s[/math], it hit [math]5 m/s[/math]. Because we’re assuming constant acceleration, we can conclude that the motorcycle hit its average velocity for a given time window at exactly the midpoint in time. Hence, it hit [math]10 m/s[/math] after 1 second (the midpoint of the first 2-second window), and it hit [math]20 m/s[/math] after 3 seconds (the midpoint of the next 2-second window). Since it hit [math]5 m/s[/math] 1 second before it hit [math]10 m/s[/math] (which it hit after 1 second), [math]5 m/s[/math] was its starting velocity.
Now, knowing the motorcycle’s starting velocity and (constant) acceleration, all we have to do is add the acceleration to its initial velocity for each passing second. After 1 second, it’s going [math]5 m/s + 5 m/s/s * 1 sec = 10 m/s[/math], which we already knew. After 5 seconds, it’s going [math]5 m/s + 5 m/s/s * 5 sec = 30 m/s[/math].
Calculus approach:
If we assume constant acceleration, we can start by representing acceleration as a function of time (t):
(1) [math]a(t) = A[/math], where A is the acceleration constant
Taking the indefinite integral of acceleration with respect to time gives us velocity:
(2) [math]v(t) = At + Vo[/math], where [math]Vo[/math] is the constant normally called [math]C[/math] in beginning calculus books, which physically represents the initial velocity in this context
Taking the indefinite integral of velocity with respect to time gives us position:
(3) [math]x(t) = (1/2)At^2 + (Vo)t + Xo[/math], where [math]Xo[/math] is the new “[math]C[/math]” representing initial position. We can place our origin anywhere, so let’s keep it simple and say [math]Xo = 0[/math].
The info given in the original question, translated to math, tells us [math]x(2) = 20[/math] and [math]x(2+2) = 20 + 40[/math]. Substituting these data pairs into equation (3) gives us this 2x2 system:
[math]x(2) = (1/2)A(2)^2 + Vo(2) + 0 = 20[/math]
[math]x(4) = (1/2)A(4)^2 + Vo(4) + 0 = 20 + 40[/math]
Solving the above system yields [math]A = 5 m/s/s[/math] and [math]Vo = 5 m/s[/math]. Now just substitute these values into equation (2):
[math]v(5) = (5)(5) + (5) = 30 m/s[/math].
It has a simple answer!
The total distance travelled is 80 km.
Half distance (40 km) is travelled at 80 km/h and the remaining half distance (40km) is travelled at 40 km/h.
Time used:
Formula : t=d/v
- t=40/80=0.5 hour or 1/2 hour
- t=40/40=1 hour
Total time taken = 1.5 hour
Now,
Speed = distance/time = 80/1.5 = 53.33 km/h
Answer : 53.33 km/h
Let the total distance be 100 km.
Average speed = total dist covered/time taken
= 100 / (30/20+60/40+10/10)
= 100 / (3/2+3/2+1)
= 100 / (3+3+2)/2
= 100×2 / 8
= 25 km/hr
Alternate method
10% of journey's = 40 km
Then, total journey = 400 kms
And, Average speed = Total dist/Total time
30% of journey
= 400 × 30/100
= 120 km
60% of journey
= 40...
Let the total distance be 100 km.
Average speed = total dist covered/time taken
= 100 / (30/20+60/40+10/10)
= 100 / (3/2+3/2+1)
= 100 / (3+3+2)/2
= 100×2 / 8
= 25 km/hr
Alternate method
10% of journey's = 40 km
Then, total journey = 400 kms
And, Average speed = Total dist/Total time
30% of journey
= 400 × 30/100
= 120 km
60% of journey
= 40...
There is an ambiguity in the question.
If you mean one third of the journey by time.
Assume the time is 12h
The first 1/3 of the journey or 4h you would travel 80 km
The second 1/4 of the journey or 3h you would travel 90 km
The rest of the journey 5h you would travel 250 km
In 12h you would travel 420 km for an average speed of 35 km/h
If you mean one third of the journey by distance.
Assume the distance is 300 km
The 1/3 of the journey 100 km would have taken 5h at 20km/h
The next 1/4 of the journey 75 km would have taken 2.5h at 30km/h
The rest of the journey 125 km would have taken 2.5h at 50 km/h
In
There is an ambiguity in the question.
If you mean one third of the journey by time.
Assume the time is 12h
The first 1/3 of the journey or 4h you would travel 80 km
The second 1/4 of the journey or 3h you would travel 90 km
The rest of the journey 5h you would travel 250 km
In 12h you would travel 420 km for an average speed of 35 km/h
If you mean one third of the journey by distance.
Assume the distance is 300 km
The 1/3 of the journey 100 km would have taken 5h at 20km/h
The next 1/4 of the journey 75 km would have taken 2.5h at 30km/h
The rest of the journey 125 km would have taken 2.5h at 50 km/h
In 10h you would travel 300 km for an average speed of 30km/h
Please answer this, I really need it right now.
Max: The formulation of your question is confusing. In the first sentence you give the speed of the motorcycle as 60 Km. Is this 60Km/hr? If it is, you have already answered your question. If it it is not the speed, but the distance, the question would make sense, but the first sentence should be “A motocyclist traveled 60 kms at a constant speed”. In that case, Luke Asswalker answer could be the right one. But how did he reach the conclusion that the motorcycle speed was 20 Km/hr? That is what you have to work out. Ranjan Karnad interpreted the first sentence as giving the speed of the motorc
Max: The formulation of your question is confusing. In the first sentence you give the speed of the motorcycle as 60 Km. Is this 60Km/hr? If it is, you have already answered your question. If it it is not the speed, but the distance, the question would make sense, but the first sentence should be “A motocyclist traveled 60 kms at a constant speed”. In that case, Luke Asswalker answer could be the right one. But how did he reach the conclusion that the motorcycle speed was 20 Km/hr? That is what you have to work out. Ranjan Karnad interpreted the first sentence as giving the speed of the motorcycle, and assumed the question was : “which was the distance covered and how long did he take to cover it?”
Let, the total distance covered by the man is ‘D’ kms.
T1 = (30D/100÷20) hours = 3D/200 hours.
T2 = (60D/100÷40) hours = 3D/200 hours
And, T3 = (10D/100÷10) hours = 2D/200 hours.
T = T1 + T2 + T3 (‘T’ be the total time taken)
Now, V(av) = D/T = D/(T1 + T2 + T3) = D/(3D/200 + 3D/200 + 2D/200) = D/D/25 = 25
The average velocity of the man is 25 kms/hr. (answer)
Time taken to cover the first 10 km =10/10 =1 hr
Time taken for the second 10 km = 10/20 = 1/2 hr
Time taken for the third 10 km = 10/30 =1/3 hr
Time taken for the last 10km = 10/40 = 1/4 hr
Total time taken to cover 40 km = 1+1/2+1/3+1/4 =(12+6+4+3)/12 =25/12 hrs
Average speed = 40/(25/12) = 96/5 =19.2 km/hr
Let us find what portion the last part of the journey is:
1/4+1/3+n=1
3/12+4/12+n=1
7/12+n=1
n=5/12
So let me restate the problem: 1/4 of the distance travelled is at 10 km/hr. 1/3 of the distance travelled is at 20km/hr. Finally 7/12 of the distance travelled is at 30km /hr.
distance = velocity * time
d=vt
The answer will be the same regardless of the total distance travelled. I will arbitrarily choose 12km as the distance. (actually I chose 12 because each fraction has 12 as a denominator.) 1/4 d=3 km. 1/3d=4km. 5/12d=5km
3km=10km/hr*t
t=3/10hr
4km=20km/hr*/t
t=1/5hr
5 km=30km/hr*t
t=1/6 hour
3/10hr+1/5hr+
Let us find what portion the last part of the journey is:
1/4+1/3+n=1
3/12+4/12+n=1
7/12+n=1
n=5/12
So let me restate the problem: 1/4 of the distance travelled is at 10 km/hr. 1/3 of the distance travelled is at 20km/hr. Finally 7/12 of the distance travelled is at 30km /hr.
distance = velocity * time
d=vt
The answer will be the same regardless of the total distance travelled. I will arbitrarily choose 12km as the distance. (actually I chose 12 because each fraction has 12 as a denominator.) 1/4 d=3 km. 1/3d=4km. 5/12d=5km
3km=10km/hr*t
t=3/10hr
4km=20km/hr*/t
t=1/5hr
5 km=30km/hr*t
t=1/6 hour
3/10hr+1/5hr+1/6hr = 9/30hr+6/30hr+5/30hr = 2/3hr.
So the journey took 2/3 hour to go 12 km
12=v*2/3hr
v=18km/hr
The average speed is 18 km/hr
Suppose that Total Distance covered by the object is X
1/3rd Distance is covered at 40kmph and 1/4th is covered at 30kmph and rest is covered at 50kmph
X/3 = 40
X/4 = 30
5X/12= 50
{ X -( X/3 + X/4 ) = X - ( 7X/12 ) = 5X/12 }
We Know Distance Avg. Speed = (d1+d2+d3 ….) / (t1+t2+t3+….)
[here ‘t’ is d/v]
{t1 = d1/ v1 = X/3 / 40 = X/120}
{t2 = X/4 / 30 = X/120}
{t3 = 5X/12 / 50 = 5X/600}
[t1+t2+t3 = 15X/600]
[d1+d2+d3 = X/3 + X/4 + 5X/12 = 12X/12 = X]
Avg. Speed = X / 15X/600
= 600X / 15X
= 40 kms
Hello,
The general method of finding average speed (not velocity) is just divide total distance travelled by total time taken.
Mathematically,
- Av. Speed = (total distance travelled÷total time taken)
Now coming to this question,
Time taken in first 40km=(40÷80) =1/2 hour
Time taken to travel next 40km=(40÷40) =1 hour
Now total time taken to travel 80km= 1/2+1 = 1.5hours,
According to formula mentioned above,
Av. Speed = 80/1.5 = 53.33km/h.
Hope that it is clear.
Let us understand this problem with a pictorial figure.Delhi is represented by point A and Agra is represented by point B.Let,S be the distance between A to B.X be the average speed of the man while going from A to B.Y be the average speed of the man while going from B to A.t1 be the time taken during A to B.t2 be the time taken during B to A.
We know,average speed = Total Distance/ Total
Time taken
[math]=(S+S)/(t1+t2)[/math]
[math]t1=S/X ;t2=S/Y[/math]
(since, Time=Distance/Speed )
average speed [math]=2S/(S/X+S/Y)[/math]
[math]=2S/S(1/X+1/Y)[/math]
[math]=2XY/(X+Y)[/math]
Therfore,average speed [math]=2XY/(X+Y)[/math]
[math]=2×40×24/(40+24)[/math]
[math]=30 [/math]kmn/hr
Let us understand this problem with a pictorial figure.Delhi is represented by point A and Agra is represented by point B.Let,S be the distance between A to B.X be the average speed of the man while going from A to B.Y be the average speed of the man while going from B to A.t1 be the time taken during A to B.t2 be the time taken during B to A.
We know,average speed = Total Distance/ Total
Time taken
[math]=(S+S)/(t1+t2)[/math]
[math]t1=S/X ;t2=S/Y[/math]
(since, Time=Distance/Speed )
average speed [math]=2S/(S/X+S/Y)[/math]
[math]=2S/S(1/X+1/Y)[/math]
[math]=2XY/(X+Y)[/math]
Therfore,average speed [math]=2XY/(X+Y)[/math]
[math]=2×40×24/(40+24)[/math]
[math]=30 [/math]kmn/hr
Data:
Vi = 40 ms-1 (velocity initial)
a = 8 ms-2 (acceleration)
t = 10s (time)
s = ? . (distance)
D = ?
equation:
vi(t)+1/2+{a)(t)
S= 40ms-1(10s)+1/2(8ms-2)(10s)2
S= 40×10+1/2×8×100
S= 400+1/2+800
S= 400+400
S= 800m Ans.
Mathematically we use Harmonic Mean to calculate the average speed, but I assume the concept of Harmonic Mean doesn't exist, let's calculate it with pure logic.
The total distance covered by car is 80 KM(40+40).
Now their are two cases:
Case 1) Distance covered 40km, speed 80km/h. Now we all know speed=Distance/ time
So Time taken is 0.5 hour.
Case 2) Now again distance covered is 40 km but speed is 40Km/h
So time taken is 1hrs
Now total distance traveled is 80KM. Total time taken is 1.5 hrs(0.5+1)
Average speed =Total Distance/Total Time
Average Speed = 80/1.5
Which is equal to 53.33km/h
Let distance covered =d
1/2 distance =d/2
Speed for half journey=40kmps
Time =distance/speed
Thus
Time =d/80
Speed of second half journey =d/2/60
=d/120
Total time=d/80 +d/120
Thus total time =(3d+2d)/240
=d/48
Speed = distance /time
=d/(d/48)
=48
Thus average speed =48kmps
Giving you your homework answer will not help you. Plus, the answer itself is completely useless, the important part is the thinking that goes in to understanding the problem and how to break it down into manageable pieces.
Twenty years from now, no one will even remember a homework question you had back in 2018. But your bosses, colleagues and any subordinates are going assess you on a range of things. One of these will be your ability to understand the problems you face at work, and how to break them down into manageable pieces.
Oh, and at home your ability to do the same will impact your pers
Giving you your homework answer will not help you. Plus, the answer itself is completely useless, the important part is the thinking that goes in to understanding the problem and how to break it down into manageable pieces.
Twenty years from now, no one will even remember a homework question you had back in 2018. But your bosses, colleagues and any subordinates are going assess you on a range of things. One of these will be your ability to understand the problems you face at work, and how to break them down into manageable pieces.
Oh, and at home your ability to do the same will impact your personal relationships.
So do not skip the process, as that is the real goal of the question. It's not the answer, rather it is the process that is treasure you get to take with you.
Best wishes.
Any distance in the set [math]x\in (0,c*10)[/math] is a possible answer. To be more specific, you have to add at least one more condition, such as with constant acceleration. You might also want to specify that the motorcycle is using its own engine and is not in a different inertial reference frame.
These aren’t trivial additions. Motorcycles generally do not produce constant acceleration — they are more likely to produce constant power when you crank them up, and then there is the shifting of gears. In intro physics classes, of course, one usually just ignores all of this to get the student to focus on “s
Any distance in the set [math]x\in (0,c*10)[/math] is a possible answer. To be more specific, you have to add at least one more condition, such as with constant acceleration. You might also want to specify that the motorcycle is using its own engine and is not in a different inertial reference frame.
These aren’t trivial additions. Motorcycles generally do not produce constant acceleration — they are more likely to produce constant power when you crank them up, and then there is the shifting of gears. In intro physics classes, of course, one usually just ignores all of this to get the student to focus on “simple” constant acceleration kinematics. Guessing that I’m doing your physics homework for you (tut! tut! for shame! do it yourself!) I’ll give the very simplest answer for constant acceleration:
[math]a = \frac{20-8}{10} = 1.2[/math] m/sec[math]^2[/math].
[math]\Delta x = \frac{v_f^2 - v_0^2}{2 a}[/math]
(kinematic relation found by solving equations of motion and eliminating time.) Plugging in:
[math]\Delta x = \frac{400 - 64}{2.4} = 336/2.4 = 140[/math] meters
(assuming I didn’t make any trivial arithmetic errors, always possible).
For grins, suppose we have constant power instead and no gear shifts or friction or drag and that we know the mass [math]m[/math] of the motorcycle (which obviously matters). Then algebraically:
[math]P_0 = \frac{mv_f^2 - mv_0^2}{2\Delta t} = Fv[/math]
throughout the motion. (This is the total work done divided by the total time for constant power, which is also the force times the speed in the direction of the force.) [math]P_0[/math] is a computable number (if we know the mass), so we can solve for the force:
[math]F = \frac{P_0}{v} = m a = m \frac{dv}{dt}[/math]
This is a very different result than we get for constant acceleration, and yet is entirely plausible — a lot more plausible than a motorcycle that delivers a constant forward directed force, actually! Now we have to solve for [math]v(t)[/math] (and it turns out that the pesky [math]m[/math] doesn’t matter after all).
[math]v dv = \frac{P_0}{m} dt[/math]
[math]\int_{v_0}^v v dv = \frac{1}{2}(v^2 - v_0^2) = \frac{P_0}{m}\int_0^t dt = \frac{P_0 t}{m}[/math]
This yields:
[math]\frac{dx}{dt} = v(t) = \sqrt{v_0^2 + \frac{2P_0 t}{m}}[/math]
which can be integrated, with a tiny bit of pain, to find [math]\Delta x.[/math]
I’d do the integral, but I’m out of time, so consider it an exercise to make up for the fact that I did your homework for you.
Average speed is defined as “total distance / total time."
let total distance be 2x
time taken in first half: t1 = x/60 hr
time taken in second half: t2 = x/ 40 hr
Average speed = total distance/total time = 2x/ {(x/60) + (x/40)} = 120(2x)/(2x+3x) =48km/hr
ANS : 48 Km/hr
Calculate the time separately
1st 10 km time is 10/10=1h
2nd 10km time is 10/20=0.5h
3rd 10km time is 10/30=0.33h
4th 10km time is 10/40=0.25h
Total Time Taken is 2.08333333h
Avg Speed is distance/time=40/2.083333=19.2km/h
Ans. 19.2 km/h
20x5280 = 105,600 feet
40x1.466 = 58.64 ( feet per second)
40 mph
105,600÷ 58.64 =1800.8185 (seconds)
1800 ÷ 60 = 30 ( minutes)
60mph your doing 88 ft per second.
105,600÷88= 1200 (seconds)
1200÷ 60 = 20 ( minutes)
total time is 50 minutes
40 miles is 211,200 feet
211,200÷ 3,000 (seconds)= 70.4 ( ft per second average)
70.4÷ 1.466= 48.02 mph
average speed is 48 mph
The definition of average speed = total distance / total time.
i.e total time = total distance / average speed = 130/ 65 = 2 hours.
So now you know how long the trip SHOULD take. How much time has already been used?
time so far = distance / speed = 90 / 60 = 1.5 hours
What time is left? 2–1.5 = 0.5 hours
Is it possible to do this? yes. There is still some time left.
What speed is needed ? Speed = distance / time = 40 / 0.5 = 80 km/hr
At this speed the entire trip takes 2 hours for 130 km i.e. 65 km/hr
Let's break it down.
Distance is speed * time.
And for simplicity let's assume change in speed is immediate.
So 30 kmph for 10 minutes is 5 km of distance.
60 kmph for 20 minutes is 20 km of distance
40 kmph for 30 minutes is 20 km of distance.
Combined you have 45km of distance.
Combined you have 10+20+30 minutes…exactly one hour. Convenient.
So the result is 45kmph.
Hope the teacher finds out you asked on Quora :)
10 km at 30 kph takes 20 minutes
40 km at 50 kph takes 48 minutes
Therefore
50 km has taken 68 minutes
Average speed is 50/68 km per minute = 0.7353 km per minute
Or multiply by 60 minutes to get 44.12 kph.