# Thread: [SOLVED] Mechanics Calculus - Average Speed

1. ## [SOLVED] Mechanics Calculus - Average Speed

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.

2. 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

3. 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!