Particle Position & Velocity: Solving s=5i+4r^2j

[scrubber]

Homework Statement

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.

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

 

[Sunil Simha]

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

Hi,

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

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

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

 

[scrubber]

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.

Thanks for your reply.

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?

 

[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}.

 

[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).

 

[dauto]

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