The x and y coordinates of a 3.70-kg particle moving in the xy plane under the influence of a net force F are given by

x = t4 − 5t
and
y = 7t2 + 1,
with x and y in meters and t in seconds. What is the magnitude of the force F at
t = 4.20 s?


N

Could some body show me how to get to the answer key 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?

You have to find the 2nd derivative of position (acceleration) and multiply that by the mass to find the x and y components, then take the sqrt of x^2+y^2 to find the total force.

To find the magnitude of the force F at t = 4.20 s, we need to differentiate the equations for x and y twice with respect to time (t) to find the acceleration components (a), and then we can use Newton's second law of motion (F = ma) to calculate the force.

Let's start by finding the acceleration components of the particle:

Given:
x = t^4 − 5t
y = 7t^2 + 1

Differentiating x twice with respect to t, we get:
a_x = d^2x / dt^2
= d / dt (4t^3 - 5)
= 12t^2

Differentiating y twice with respect to t, we get:
a_y = d^2y / dt^2
= d / dt (14t)
= 14

Now, we can calculate the acceleration vector a at t = 4.20 s:

Substituting t = 4.20 s into the acceleration components, we get:
a_x = 12(4.20)^2
a_x ≈ 212.16 m/s^2

a_y = 14

To calculate the magnitude of the force F, we use Newton's second law of motion:

F = ma

Substituting the acceleration components, we get:
F = √(a_x^2 + a_y^2)
= √(212.16^2 + 14^2)
≈ √(44960.2056 + 196)
≈ √(45156.2056)
≈ 212.6 N

Therefore, the magnitude of the force F at t = 4.20 s is approximately 212.6 N.

forcex=t^4-5t=4.20^4-5(4.20)=290

forcey=7t^2+1=7*(4.90)^2 +1=169

force=sqrt(forcex^2+forcey^2)=336 N