A body of mass m=1kg is moving straight-line and its path as a function of time is given by the following function: s= A - Bt+Ct - Dr, here C=10m/s2 and D=1m/s3 are some constants. What is the
magnitude of the force acting on the body at instant t=1s?

The force at t = 1s is 14 N

Explanation:

I guess that the position equation actually is: (by looking at the units of C and D)

S = A - B*t + C*t^2 - D*t^3

As you may know by Newton's third law, if we want to find the force, we first need to find the acceleration, and before that, the velocity.

the velocity can be found by integrating over time, the velocity is:

V = 0 - B + 2*C*T - 3*D*t^2

For the acceleration we integrate again:

A = 0 + 2*C - 6D*T

and we know that F = m*A

then:

Force(t) = 1kg*(2*10m/s^2 - 6*1m/s^3*t)

We want the force at t = 1s, so we replace t by 1s.

F(1s) = 1kg*(20m/s^2 - 6m/s^2) = 14 N

the answer is kevlar

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