Question 76c41

Jan 10, 2017

$\text{90 m}$

Explanation:

Your goal here is to find the acceleration of the car so that you can use the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{d = {v}_{0} \cdot t + \frac{1}{2} \cdot a \cdot {t}^{2}}}}$

Here

• $d$ is the distance traveled in the $\text{3 s}$
• ${v}_{0}$ is the initial velocity of the car
• $t$ is the time given
• $a$ is the acceleration of the car

Now, you know that an object's acceleration tells you how said object's velocity is changing with respect to time.

$a = \frac{\Delta v}{\Delta t} \textcolor{w h i t e}{\textcolor{b l u e}{\leftarrow \text{ change in velocity")/(color(purple)(larr" change in time}}}$

In your case, you know that the velocity of the car changes by

$\Delta v = {\text{25 m s"^(-1) - "15 m s}}^{- 1}$

$\Delta v = {\text{10 m s}}^{- 1}$

and that it takes $\text{3 s}$ for this change to occur. If you take $t = 0$ to be the time when the car begins to accelerate, you can say that the change in time is equal to

$\Delta t = \text{3 s"- "0 s}$

$\Delta t = \text{3 s}$

This means that the car's acceleration is equal to

$a = {\text{10 m s"^(-1)/"3 s" = 10/3 color(white)(.)"m s}}^{- 2}$

This means that with every passing second, the velocity of the car increases by $\frac{10}{3} \textcolor{w h i t e}{.} {\text{m s}}^{- 1}$.

Plug this into the first equation and calculate $d$

d = "15 m" color(red)(cancel(color(black)("s"^(-1)))) * 3color(red)(cancel(color(black)("s"))) + 1/2 * 10/3color(white)(.)"m" color(red)(cancel(color(black)("s"^(-2)))) * 3^2 color(red)(cancel(color(black)("s"^(2))))

d = "45 m" + "45 m" = color(darkgreen)(ul(color(black)("90 m")))#

Therefore, you can say that the car travels $\text{90 m}$ in the $\text{3 s}$ it's being accelerated.