Question

A particle moves along a straight line with velocity given by v(t)=7-(1.01)-t^2 at time t>0. What is the acceleration of the particle at time t=3 ?

Answer 1

The acceleration of the particle at time
`\(t = 3\) is \(-6 \ \text{m/s}^2\)`.

Step-by-step explanation:

The velocity of the particle is given by
`\(v(t) = 7 - 1.01t^2\)`. To find the acceleration, we need to take the derivative of the velocity function with respect to time (t), which gives us the acceleration function. The derivative of (v(t)) is (a(t) = -2.02t).

Now, plug in (t = 3) into the acceleration function to find the acceleration at that specific time:

`\[a(3) = -2.02 * 3 = -6.06 \ \text{m/s}^2.\]`

Therefore, at \(t = 3\), the acceleration of the particle is \(-6.06 \ \text{m/s}^2\). The negative sign indicates that the particle is experiencing deceleration at this particular moment.

It means the particle's velocity is decreasing, and the magnitude of this decrease is
`\(6.06 \ \text{m/s}^2\)`. This information is crucial in understanding the dynamics of the particle's motion along the straight line at \(t = 3\).

Answer 2

Refer to the attached image.