Question

A go-cart travels the first half of a 100m track with a constant speed of 5.00m/s. In the second half of the track, it experiences a mechanical problem and slows down at 0.200m/s^2. How long does it take the go cart to travel the 100m distance. Explain why two different answers.

Answer 1

Total time taken to travel 100 m = 23.82 seconds

In first case the velocity is constant and in the second case velocity is reducing by 0.2 m/s each second, so time taken will be different.

Step-by-step explanation:

A go-cart travels the first half of a 100m track with a constant speed of 5.00m/s

Time taken for first half of 100m = Distance velocity = 50/5 = 10 seconds

We have equation of motion
`s=ut+(1)/(2) at^2`, where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

For the second half of 100m we have

s = 50 m, u = 5 m/s, a =
`-0.200m/s^2`

Substituting

`50=5t-(1)/(2) *0.2*t^2\\ \\ 0.1t^2-5t+50=0\\ \\ t^2-50t+500=0`

t = 13.82 seconds or t = 36.18 seconds

So it will take 13.82 seconds to travel second 50m

Total time taken = 10 + 13.82 = 23.82 seconds

In first case the velocity is constant and in the second case velocity is reducing by 0.2 m/s each second, so time taken will be different.