Question

could somebody shows me how to get to the answer key. The answer key is 1290N.The x and y coordinates of a 4.30-kg particle moving in the xy plane under the influence of a net force F are given by
x = t4 − 5t
and
y = 5t2 + 1,
with x and y in meters and t in seconds. What is the magnitude of the force F at
t = 5.00 s?

Answer 1

x = t^4 - 5 t
dx/dt = 4 t^3 - 5
d^2x/dt^2 = 12 t^2 = ax

y = 5 t^2 + 1
dy/dt = 10 t
d^2y/dt^2 = 10 = ay

|a| = sqrt(100 + 144t^2)
at t = 5
|a| = sqrt (3700)
|a| = 60.8

|F| = m |a| = 4.3 * 60.8 = 261.5 N
sorry, can not get to your answer key