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

a = -47.93 m/s²

Step-by-step 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 = (-42-97)/(2.9)`

`a = (-139)/(2.9)`

a = -47.93 m/s²

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