Question

An automobile accelerates uniformly from a speed of 60 km/hr (16.67 m/s) to a speed of 80 km/hr (22.2 m/s) in 5 s. Determine the acceleration of the car and the distance traveled in this 5 sec interval.

Answer 1

Answer:

The acceleration is $1.106\ m/s^2$ and the distance covered is 97.17 m.

Explanation:

Given that,

Initial speed of an automobile, u = 60 km/hr = 16.67 m/s

Final speed of an automobile, v = 80 km/hr = 22.2 m/s

Time, t = 5 s

We need to find the acceleration of the car and the distance traveled in this 5 sec interval.  Let a is the acceleration. Using the definition of acceleration as :

$a=\dfrac{v-u}{t}\\\\a=\dfrac{22.2-16.67}{5}\\\\a=1.106\ m/s^2$

Let d is the distance covered. Using the third equation of motion to find it as follows :

$v^2-u^2=2as\\\\s=\dfrac{v^2-u^2}{2a}\\\\s=\dfrac{(22.2)^2-(16.67)^2}{2\times 1.106}\\\\s=97.17\ m$

So, the acceleration is $1.106\ m/s^2$ and the distance covered is 97.17 m.