Question

a car starts from rest and accelerates at a constant rate over a time of 18.89 seconds, for a distance of 13 m. what is the acceleration of the car? 71.9 m/s^2 0.1498 14.99 m/s^2 0.729 m/s^2

Explanation:

Step1: Identify the kinematic - equation

The car starts from rest ($u = 0$ m/s), the kinematic equation $s=ut+\frac{1}{2}at^{2}$ is used. Since $u = 0$, the equation simplifies to $s=\frac{1}{2}at^{2}$.

Step2: Solve for acceleration $a$

We can re - arrange the equation $s=\frac{1}{2}at^{2}$ to get $a=\frac{2s}{t^{2}}$. Given $s = 13$ m and $t=18.89$ s.
Substitute the values: $a=\frac{2\times13}{(18.89)^{2}}$.
First, calculate $(18.89)^{2}=18.89\times18.89 = 356.8321$.
Then, $2\times13 = 26$.
So, $a=\frac{26}{356.8321}\approx0.0729$ m/s².

Answer:

$0.0729$ m/s²