Question

a particle moving along the x axis has a position given by where 22 m/s, 3.8 m/s3 and is measured in seconds. what is the magnitude of the acceleration of the particle at the instant when its velocity is zero? please give your answer in units of m/s2.

Answer 1

Therefore, the magnitude of the acceleration of the particle at the instant when its velocity is zero is approximately 18.258 m/s².

Step-by-step explanation:

To find the magnitude of acceleration at the instant when velocity is zero, we need to differentiate the given velocity equation with respect to time (t) to obtain the acceleration equation.

Given:

Velocity equation: v(t) = 22 - 3.8t^2

Differentiating the velocity equation with respect to time, we get:

a(t) = d(v(t))/dt = -2 * 3.8t

To find the magnitude of acceleration at the instant when velocity is zero, we need to solve for t when v(t) = 0.

0 = 22 - 3.8t^2

3.8t^2 = 22

t^2 = 22/3.8

t^2 ≈ 5.789

Taking the square root of both sides, we find:

t ≈ √(5.789)

t ≈ 2.403

Now we can substitute this value of t into the acceleration equation to find the magnitude of acceleration at that instant:

a(t) = -2 * 3.8 * 2.403

a(t) ≈ -18.258

Therefore, the magnitude of the acceleration of the particle at the instant when its velocity is zero is approximately 18.258 m/s².