Question

A shuffleboard disc of mass 0.50 kg accelerates under an applied force of 12.0 N [forward]. if the disk moves from rest for 0.20s, how far does it travel while accelerating?
And also what would the final velocity be?

Answer 1

First you must find acceleration

\(F=ma\)
F=12.0 N
m=0.5 kg
hence, a=12/0.5 = 24.0 m/s^2

Now, I'd like to solve for the second question first, which is finding final velocity.
\(v=\int_{ }^{ }a\ dt\)
\(v=\int_{ }^{ }24\ dt\ =\ 24t+v_{o}\)
Since the disk moves from REST, the initial velocity is 0, and hence the velocity (final velocity) can be represented as \(v=24t\). It moves for 0.20 seconds, so plugging that in, \(0.20\cdot24\) = \(4.8~m/s\)

Now back to the first question: how far does it travel while accelerating:
\(x=\int_{ }^{ }\left|v\left(t\right)\left|dt=\int_{ }^{ }\right|24t\right|dt=12t^{2}\)
t=0.2
hence distance is \(0.2^{2}\cdot12=\)
You can calculate that

Answer 2

You can also use kinematics equations here

for question 2 -
\(a\approx\frac{v_{f}-v_{i}}{t}\)

For question 1 -
\(x=x_{i}+v_{i}t+\frac{1}{2}at^{2}\)