Question

A bicyclist starts from rest and accelerates at a rate of 2.3 m/s^2 until it reaches a speed of 23 m/s. It then slows down at a constant rate of 1.0 m/s^2 until it stops. How much time elapses from start to stop? We assume an answer in seconds

Answer 1

33 seconds.

Step-by-step explanation:

The equation for speed with constant acceleration at time t its:

`V(t) \ = \ V_0 \ + \ a \ t`

where
`V_0` is the initial speed, and a its the acceleration.

First half of the problem

Starting at rest, the initial speed will be zero, so

`V_0 = 0`

the final speed is

`V(t_(f1)) = 23 (m)/(s)`

and the acceleration is

`a = 2.3 (m)/(s^2)`.

Taking all this together, we got

`V(t_(f1)) = 23 (m)/(s) = 0 + 2.3 \ (m)/(s^2) t_(f1)`

`23 (m)/(s) = 2.3 \ (m)/(s^2) t_(f1)`

`(23 (m)/(s))/(2.3 \ (m)/(s^2)) =  t_(f1)`

`10 s =  t_(f1)`

So, for the first half of the problem we got a time of 10 seconds.

Second half of the problem

Now, the initial speed will be

`V_0 = 23 (m)/(s)`,

the acceleration

`a=-1.0 (m)/(s^2)`,

with a minus sign cause its slowing down, the final speed will be

`V(t_(f2)) = 0`

Taking all together:

`V(t_(f2)) = 0 = 23 (m)/(s) -  1.0 (m)/(s^2) t_(f2)`

`23 (m)/(s) =  1.0 (m)/(s^2) t_(f2)`

`(23 (m)/(s))/(1.0 (m)/(s^2)) = t_(f2)`

`23 s = t_(f2)`

So, for the first half of the problem we got a time of 23 seconds.

Total time

`t_total = t_(f1) + t_(f2) = 33  \ s`