Question

At a certain time, a particle had a speed of 97 m/s in the positive x-direction, and 2.9 s later its speed was 42 m/s in the opposite direction. What was the average acceleration of the particle during this 2.9 s interval?

Answer 1

Answer:

 a = -47.93 m/s²

Explanation:

given,

speed of particle in positive x-direction, u = 97 m/s

speed of particle in opposite direction, v = -42 m/s

time = 2.9 s

average acceleration, a = ?

now,

$a = \dfrac{-42-97}{2.9}$

$a = \dfrac{-139}{2.9}$

   a = -47.93 m/s²

Hence, the average acceleration is equal to  -47.93 m/s²