# A car moving with constant acceleration covered the distance between two points 52.4 m apart in 5.96 s. Its speed as it passes the second point was 14.6 m/s. (a) At what prior distance from the first point was the car at rest?

Sep 9, 2015

The car was at rest $\text{2.29 m}$ before the first point.

#### Explanation:

The idea here is that once the car starts moving, its acceleration will be constant.

This means that you can use the speed with which it passes the first point and its acceleration to write

${v}_{1}^{2} = {\underbrace{{v}_{0}^{2}}}_{\textcolor{b l u e}{= 0}} + 2 \cdot a \cdot d \text{ }$, where

${v}_{1}$ - the speed it has at the moment it passes the first point;
${v}_{0}$ - the initial velocity, equal to zero since the car's starting from rest;
$d$ the disance before it passes the first point.

The values you have for ${v}_{1}$ and $a$ are correct, but I'll show how to get them for other students interested in how this problem can be solved.

You basically work with two equations with two unknowns, ${v}_{1}$ and $a$.

First, you know that

$s = {v}_{1} \cdot t + \frac{1}{2} \cdot a \cdot {t}^{2} \text{ }$, where

$s$ - the distance between the two points;
$t$ - the time needed to go from the first point to the second point;

SInce you know that car's speed when it reaches the second point, you can write

${v}_{2} = {v}_{1} + a \cdot t \implies {v}_{1} = {v}_{2} - a \cdot t$

Plug this back into the above equation to get

$s = \left({v}_{2} - a t\right) \cdot t + \frac{1}{2} \cdot a \cdot {t}^{2}$

Plug in your values to get - I'll skip the units for simplicity

52.4 = (14.6 - 5.96 * a) * 5.96 + 1/2 * a * 5.96""^2

This will eventually get you

$52.4 = 87.016 - 35.52 a + 17.76 a$

$17.76 a = 34.616 \implies a = \frac{34.616}{17.76} = {\text{1.95 m/s}}^{2}$

The speed of the car at the time it passes the first point will be

${v}_{1} = 14.6 - 1.95 \cdot 5.96 = \text{2.99 m/s}$

Now, you know that the car started from rest and reached a speed of $\text{2.99 m/s}$ by the time it passed the first point. This means that you have

${v}_{1}^{2} = 2 \cdot a \cdot d \implies d = {v}_{1}^{2} / \left(2 \cdot a\right) = \left({2.99}^{2} \text{m"^color(red)(cancel(color(black)(2)))/color(red)(cancel(color(black)("s"^2))))/(2 * 1.95color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"^2)))) = color(green)("2.29 m}\right)$

Alternatively, you can first find the time it took for the car to reach the first position

v_1 = underbrace(v_0)_(color(blue)(=0)) + a * t_b implies t_b = v_1/a = (2.99color(red)(cancel(color(black)("m")))/color(red)(cancel(color(black)("s"))))/(1.95color(red)(cancel(color(black)("m")))/"s"^color(red)(cancel(color(black)(2)))) = "1.53 s"

The distance it covered before the first position is

d = 1/2 * a * t_b^2 = 1/2 * 1.95"m"/color(red)(cancel(color(black)("s"^2))) * 1.53""^2color(red)(cancel(color(black)("s"^2))) = color(green)("2.29 m")