[SOLVED] Mechanics Calculus - Average Speed

Sep 2009
Q) A motorcyclist starts from rest at A and travels in a straight line. For the first part of the motion, the motorcyclist’s displacement x metres from A after t seconds is given by x = 0.6t^2 − 0.004t^3.

(i) Show that the motorcyclist’s acceleration is zero when t = 50 and find the speed V ms^-1 at this time.

For t ≥ 50, the motorcyclist travels at constant speed V ms^-1.

(ii) Find the value of t for which the motorcyclist’s average speed is 27.5ms^-1

My Attempt

I've already done part (i) to get v=30ms^-1
I'm stuck at part (ii). By using calculus or any other easy method kindly show me how to solve this.

Thanks in advance!
Last edited:
May 2010
Varanasi, Uttar Pradesh-India
As we see that the velocity is positive for the first 50 seconds of journey therefore net displacement=net distance traveled.

calculating displacement for first 50 seconds from\(\displaystyle x=0.6t^2-0.004t^3\) we get 900 meters.

then assuming that it travels for t more seconds at 30 m/sec therefore displacing 30t meters.

\(\displaystyle Average speed=(Total Distance)/(Total time)\)

on solving:\(\displaystyle 27.5=(900+30t)/(50+t)\)

t=190 seconds....
so net time time is 240 seconds from the starting place to get an avearge speed of 27.5m/sec
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Sep 2009
Thanks for the quick reply! The distance for the first 50s is 1000 not 900. I think you made an error in a hurry ;)

But I've got the concept behind it. Therefore, thanks!