# Calculating velocity with force

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1. Sep 30, 2015

### mia_material_x1

1. The problem statement, all variables and given/known data
A net horizontal force F=A+Bt^3 acts on a 2kg object, where A=5N and B =2N/s^3. What is the horizontal velocity of this object 4 seconds after it starts from rest?

2. Relevant equations
F=ma

3. The attempt at a solution
F=ma=5N+(2N/s^3)t^3
(2kg)a=5N+(2N/s^3)(4s)^3 ==> a= 2.5N + t^3 = 66.5m/s^2

a=dv/dt ==> v=(2.5 N)t + (1/4 N)t^4
when t= 4s
V= (2.5)(4) +(1/4)(4)^4 = 74m/s

I used V=Vi+at when Vi=0m/s, t=4s, a=66.5m/s^2, and I got V=266m/s

Which one is correct way of solving it, and why? Please let me know. Thanks

2. Sep 30, 2015

### RUber

It seems like the first method is right.
V = Vi + at only applies when acceleration is constant .

3. Sep 30, 2015

### RUber

I mean, $v(t) = v(0) + \int_0^t a(\tau) d\tau$, which as you correctly pointed out is
$v(t) =0+ \int_0^t \frac{ F(\tau)}{m} d\tau \\ \quad = \int_0^t 2.5 + \tau^3 d\tau \\ \quad = \left. (2.5\tau + \frac14 \tau^4) \right|_0^t \\ \quad =2.5t + \frac{t^4} {4}$

Since acceleration is increasing (quickly) with time, it should be pretty clear why if you use a(4) as your constant acceleration in the equation:
v(t)=v_i+at,
the result would be so much larger than you found with the correct method.

4. Sep 30, 2015

### mia_material_x1

Yeah you are right. I just assumed the acceleration was constant that's why the result was incorrect. Thanks guys!