Question

1. a race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. determine the acceleration of the car and the distance traveled.

Explanation:

Step1: Find the acceleration

Use the formula $a=\frac{v - u}{t}$, where $v$ is final - velocity, $u$ is initial - velocity, and $t$ is time.
$a=\frac{46.1 - 18.5}{2.47}=\frac{27.6}{2.47}\approx11.2\ m/s^{2}$

Step2: Find the distance traveled

Use the formula $s=ut+\frac{1}{2}at^{2}$.
First, substitute $u = 18.5\ m/s$, $a\approx11.2\ m/s^{2}$, and $t = 2.47\ s$ into the formula.
$s=(18.5\times2.47)+\frac{1}{2}\times11.2\times(2.47)^{2}$
$=18.5\times2.47 + 5.6\times6.1009$
$=45.695+34.16504$
$\approx79.9\ m$

Answer:

Acceleration: $11.2\ m/s^{2}$, Distance: $79.9\ m$