Question

a squash ball makes contact with a squash racket and changes velocity from 15 m/s West to 25 m/s East in 0.10 s. determine the vector acceleration of the squash ball.

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Answer 1

The vector acceleration of the squash ball is 100 m/s² going to east.

Here, we are going to solve for the vector acceleration of the squash ball having a change in velocity in a very short time with directions from west to east. We know that a vector is a quantity that has magnitude and direction so, we are going to solve the magnitude of the acceleration and define its direction.

In this problem, we are going to use the average acceleration.

For the formula, we will use:

a = Δv / Δt

where

a     is the average acceleration

Δv   is the change of velocity, Δv = Vf - Vi

Δt    is the change of time, Δt = Tf - Ti or the total time covered in the

      change of velocity

For the given information

Vi = 15 m/s

Vf = 25 m/s

Δt = .10s

a = ?

Solving the problem

Let us use the formula for average acceleration and substitute the given information, we have:

a = Δv / Δt

a = (Vf - Vi) / (Tf - Ti)

a = (25 m/s - 15 m/s) / .10s

a = 100 m/s²

Therefore, the vector acceleration of the squash ball is 100 m/s² going to east.

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Answer 2

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