# Homework Help: 1D motion

1. May 7, 2014

### scrubber

1. The problem statement, all variables and given/known data

The position s (in metres) of a particle, as a function of time t (in seconds), is described by the equation s=5i+4r^2j. i and j are perpendicular unit vectors.
a) Calculate the position of the particle at t=5.
b) Determine the particle's average velocity between t=0 and t=2.
c) Determine the particle's instantaneous velocity at t=4.

3. The attempt at a solution

a) When t=5,
s=5i+4(25)j=5i+100j
s=√10025=100m

b) When t=0,
s=5i=5m
When t=2,
s=5i+4(4)j=5i+16j=√281=16.76m

average velocity=displacement/time=(16.76-5)/(2-0)=5.9m/s

c) By differentiation,
v=4(2)tj=8tj
When t=4,
v=8(4)tj=32j=32m/s

2. May 7, 2014

### Sunil Simha

Hi,

The position of a particle is a vector right? √(52+1002) is the distance.

Again, average velocity is a vector. Displacement is the vector difference of the final and initial positions.

3. May 7, 2014

### scrubber

Do you mean I have to give the direction as well? The angles?
For a),
tanθ=100/5, θ=1.5°
Therefore, the position is 100m 1.5°

And for b), tanβ=16/5, β=1.3o
Therefore, the average velocity is 5.9m/s 1.3°

Like these?

4. May 7, 2014

### Sunil Simha

Your answer is right for part (A) but not for part (b).

In part (b), you are supposed to find $\frac{\vec{s_2}-\vec{s_1}}{Δt}$. You just have to subtract each component (i.e. like final x-initial x and final y- initial y) separately and then divide the resultant vector by t. Your answer should look like $\frac{(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}}{Δt}$.

5. May 7, 2014

### dauto

You made a mistake on part b because yo wrote
s=5i=5m
Never ever do that.
either something is a vector (5i is a vector), or something is a scalar (5m is a scalar). But something cannot be both a scalar and a vector at the same time, so the equation 5i=5m makes no sense. if s is a vector than you should denote that somehow. One common convention is to use boldface so your equation should look like this
s=5i
|s|=5m.

Always make sure to do that (even on your scraps that you're going to throw away anyways).

6. May 7, 2014

### dauto

One more thing. This is a 2-D motion but the title of the thread says 1-D motion.