# Calculating Velocity from a variable force.

## Homework Statement

A body of mass 5 kg is acted on by a force in a straight line. The magnitude of the force after t seconds is given by (2t - 3t^2) kg wt. If the body has an initial velocity of 3ms-1 in the same direction as the force, calculate its velocity after 4 seconds.

∑F = ma
a = dV/dt

## The Attempt at a Solution

So what I've done is this.

m = 5
F = (2t - 3t^2)
a = (2t - 3t^2) / 5
dV/dt = (2t - 3t^2) / 5
v = ∫(2t - 3t^2) / 5 dt
v = (t^2 - t^3) / 5 + c

Using t = 0, v = 3 (from initial velocity), solving for c yields c = 3. So,

v = (t^2 - t^3)/5 + 3 and substituting t = 4, I get v = -6.6ms-1

The back of my textbook gives an answer of -91.08ms-1, could someone please tell me where I've gone wrong?

## Answers and Replies

TSny
Homework Helper
Gold Member
Hello and welcome to PF!

I think your answer is correct. I don't see how you could get an answer of -91 m/s.

TSny
Homework Helper
Gold Member
The wording of the problem is a little odd. "Magnitude" of a force is generally defined to be a positive quantity.
Yet, the expression 2t - 3t2 takes on negative values during the time interval 2/3 < t < 4.

Hello and welcome to PF!

I think your answer is correct. I don't see how you could get an answer of -91 m/s.

I just had a look at all of the question again and realised that the units which the force was given in are kg wt. I forgot to make the conversion between kg wt and N, and that threw me out.
(2t - 3t^2) kg wt = g(2t - 3t^2) N

Using F = g(2t - 3t^2), the correct answer of -91.08m/s was obtained.

TSny
Homework Helper
Gold Member
OK. Good. I did not notice the kg wt.