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

Answer:

33 seconds.

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

$V(t_{f1}) = 23 \frac{m}{s}$

and the acceleration is

$a = 2.3 \frac{m}{s^2}$.

Taking all this together, we got

$V(t_{f1}) = 23 \frac{m}{s} = 0 + 2.3 \ \frac{m}{s^2} t_{f1}$

$23 \frac{m}{s} = 2.3 \ \frac{m}{s^2} t_{f1}$

$\frac{23 \frac{m}{s}}{2.3 \ \frac{m}{s^2}} = t_{f1}$

$10 s = t_{f1}$

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

$V_0 = 23 \frac{m}{s}$,

the acceleration

$a=-1.0 \frac{m}{s^2}$,

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

$V(t_{f2}) = 0$

Taking all together:

$V(t_{f2}) = 0 = 23 \frac{m}{s} - 1.0 \frac{m}{s^2} t_{f2}$

$23 \frac{m}{s} = 1.0 \frac{m}{s^2} t_{f2}$

$\frac{23 \frac{m}{s}}{1.0 \frac{m}{s^2}} = t_{f2}$

$23 s = t_{f2}$

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

Total time

$t_total = t_{f1} + t_{f2} = 33 \ s$