Hi guys, i need help with this, please provide explanation to how you achieved your answer.
Car A & Car B travel in the same direction for 40 seconds. Car A is traveling at a constant velocity of 40 ms-1, overtakes car B at T=0, in order to catch up with Car A, car B immediately accelerates uniformly for 20 seconds to reach a constant velocity of 50ms-1 .
1) Calculate the additional time for B to catch up with A.
2) The distance each car will have travelled since T=0.
Next to the question there is a Velocity Vs time graph. T on the X-axis going from T=0 till T=40. Velocity for A is constant at 40ms-1, Velocity for B increases from 25ms-1 to 50ms-1 from T=0 to T=20, then travels at a constant 50ms-s from T=20 to T=40 seconds. )
Any help will be thoroughly appreciated, thank you.
Answers & Comments
Verified answer
Hello
find the acceleration of the second car
a = (v2-v1)/t = (50 - 25) m/s /20s = 1.25 m/s^2
find the distance driven by car 2 during its acceleration phase:
s = v0*t + 1/2*at^2 ( with a = 1.25 m/s^2, v0 = 25 m/s)
s = 25*20 + 0.5*1.25*20^2 = 750 m
the distance driven by car 1 until car 2 overtakes it is
s = v*t = (40*t).
the distance driven by car 2 until it overtakes car 1 = 750 m + v*t = 750 m + 50*t1
with t = (20 + t1) --> t1 = t - 20
for obvious reasons the distances driven by car 1 and car 2 must be equal:
40t = 750 + 50(t - 20)
- 10t = - 250
t = 25 seconds <---- = total time needed for both cars to meet <--- ans.
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distance travelled = v*t = 40*25 = 1000 m (first car)
or 750 + 50(25 - 20) = 750 + 50*5 = 1000 m (second car) <--- ans.
Regards
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