Question

A car is traveling to the right with a speed of 29m/s ​ when the rider slams on the accelerator to pass another car. The car passes in 110m with constant acceleration and reaches a speed of 34m/s.What was the acceleration of the car as it sped up?Answer using a coordinate system where rightward is positive. Round the answer to two significant digits.

Answer 1

Answer:

Acceleration of the car is $1.43\ m/s^2$.

Explanation:

It is given that,

Initial speed of the car, u = 29 m/s

Finally it reaches a speed of, v = 34 m/s

Distance, d = 110 m

We need to find the acceleration of the car as it speed up. It can be calculated using third law of motion as :

$v^2-u^2=2ad$

$a=\dfrac{v^2-u^2}{2d}$

$a=\dfrac{(34)^2-(29)^2}{2\times 110}$

$a=1.43\ m/s^2$

So, the acceleration of the car as it speeds up is $1.43\ m/s^2$. Hence, this is the required solution.