Question

The position of a particle moving along the x-axis is given by x(t)=sin(2t2+5t) , where x(t) is in meters and t is in seconds. Find the particle’s acceleration at t = 2 seconds.

Answer 1

To find the acceleration of the particle at t = 2 seconds, we need to find the second derivative of the position function x(t).

Taking the derivative of x(t) with respect to t using the chain rule:

x'(t) = cos(2t^2 + 5t) * (4t + 5)

Taking the derivative again:

x''(t) = -sin(2t^2 + 5t) * (4t + 5)^2 + cos(2t^2 + 5t) * (8t + 5)

Now we can find the acceleration at t = 2 seconds by plugging t = 2 into x''(t):

x''(2) = -sin(2(2^2) + 5(2)) * (4(2) + 5)^2 + cos(2(2^2) + 5(2)) * (8(2) + 5)

Simplifying:

x''(2) = -sin(18) * (13)^2 + cos(18) * (16 + 5)

Using a calculator to find sin(18) ≈ 0.309 and cos(18) ≈ 0.951:

x''(2) ≈ -0.309 * 13^2 + 0.951 * 21

x''(2) ≈ -0.309 * 169 + 0.951 * 21

x''(2) ≈ -52.221 + 19.971

x''(2) ≈ -32.25

Therefore, the particle's acceleration at t = 2 seconds is approximately -32.25 m/s^2.