Question

A runner maintains constant acceleration after starting from rest as she runs a distance of 60.0 m. The runner's speed at the end of the 60.0 m is 9.00 m/s. How much time did it take the runner to complete the 60.0 m distance?

Answer 1

Answer:$\frac{40}{3}s$

Step-by-step explanation:

This is a typical problem of u.a.r.m. ( uniformly accelerated rectilinear motion). Let's take an eye on the u.a.r.m. equations:

$v(t) = v_{0} + a.t\\x(t) = x_{0} + v_{0}.t + \frac{1}{2}.a.t^{2}$

We know the runner started from rest so we can discard $v_{0}$ as $v_{0} = 0$

Also we can consider she starts at $x_{0} = 0$.

All we have to do now is replace these values at our equations to obtain:

$v(t) = a.t\\x(t) = \frac{1}{2}.a.t^{2}$

As we are told that the acceleration remains constant, we can then find the time it took for her to run the whole distance as follows:

First we note that the ratio between her velocity and the time it takes for her to run a certain distance is constant. This is:

$\frac{v(t)}{t} = a ; \forall t$

In particular, if we take $t=t_{f}$ where $t_{f}$ is the total time it takes for her to run the whole distance $x(t=t_{f})=60m$

We know that at this point the velocity is:

$v(t=t_{f}) = 9\frac{m}{s}$

So that:

$a = \frac{v(t=t_{f})}{t_{f}} = \frac{9\frac{m}{s}}{t_{f}}$

Now we can replace this in the equation of motion $x(t)$

$x(t=t_{f}) = \frac{1}{2}.a.t^{2} = \frac{1}{2}.\frac{9\frac{m}{s}}{t_{f}}.t_{f}^{2} = 60m$

Where we finally find that

$t_{f} = \frac{40}{3}s$

Answer 2

Answer:

It takes to the runner 13.33 seconds to complete the 60 m.

Step-by-step explanation:

This is a uniformly accelerated rectilinear motion problem.

$V_0 = 0  m/s (Rest)$

$V_f = 9 m/s$

$d= 60 m$

$V_f^2-V_0^2 = 2ad$

Isolating "a" from the equation:

$a=\frac{V_f^2-V_0^2}{2d} =\frac{(9 (m/s))^2-(0 (m/s))^2}{2(60 m)}  =\frac{81 m^2/s^2}{60 m} = 0.675  m/s^2$

On the other hand we have a second equation associated with uniformly accelerated motion:

$a=\frac{V_f-V_0}{t}$

Isolating "t" from this equation, We have:

$t=\frac{V_f-V_0}{a}=\frac{9 m/s -0 m/s}{0.775 m/s^2}  = 13.33 s$

So, the runner takes 13.33 seconds to complete 60 meters.