# A car, which is accelerating uniformly from rest, travels a distance of 28.5 m in the tenth second of its motion. What is its acceleration?

## The only thing I can think of is to use v=u+at, where u = 0 m/s, t = 10 s and v = 28.5 m/s. But the velocity is constantly increasing so the final velocity value is incorrect. Hmm.

Feb 14, 2018

The acceleration is ${\text{3 m/s}}^{2}$.

#### Explanation:

Let the acceleration of the car be $a$. The initial velocity will obviously be $\text{0 m/s}$ as the car was at rest. Putting this into Newton's Second Law of Motion, we get

$0 \cdot 10 + \frac{1}{2} \cdot a \cdot {10}^{2} - \left(0 \cdot 9 + \frac{1}{2} \cdot a \cdot {9}^{2}\right) = 28.5$

$\implies 50 a - 40.5 a = 28.5$

$\implies a = 3$

To find the distance traveled in the 10th second, we have to subtract the total distance traveled in $9$ seconds from the total distance traveled in $10$ seconds.