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. (1 point) Responses −124.28 m/s2 − 124 .28 m / s 2 −114.60 m/s2 − 114 .60 m / s 2 12.40 m/s2 12 .40 m / s 2 126.92 m/s2 126 .92 m / s 2 129.56 m/s2
To find the particle's acceleration at t = 2 seconds, we need to take the second derivative of the position function x(t).
First, we find the first derivative of x(t) with respect to t:
x'(t) = d/dt (sin(2t^2 + 5t))
= cos(2t^2 + 5t) * (4t + 5)
Next, we find the second derivative by taking the derivative of x'(t) with respect to t:
x''(t) = d/dt (x'(t))
= d/dt (cos(2t^2 + 5t) * (4t + 5))
= -sin(2t^2 + 5t) * (4t + 5)^2 + cos(2t^2 + 5t) * 4
Now we can substitute t = 2 into x''(t) to find the particle's acceleration at t = 2 seconds:
x''(2) = -sin(2(2)^2 + 5(2)) * (4(2) + 5)^2 + cos(2(2)^2 + 5(2)) * 4
= -sin(4 + 10) * (8 + 5)^2 + cos(4 + 10) * 4
= -sin(14) * (13)^2 + cos(14) * 4
≈ -114.60 m/s^2
Therefore, the particle's acceleration at t = 2 seconds is approximately -114.60 m/s^2.