Solving Frisbee and Student Motion: A Homework Challenge

[dramallama]

Homework Statement

Just when a Frisbee passes over the head of a student at a speed of 4m/s and decelerating at a rate of 1.5 m/s², she starts to run accelerating at a rate of 1 m/s². [neglect vertical motion of Frisbee]

a)How long will it take the student to catch the Frisbee?

b) If instead of accelerating the student were able to run at a constant speed of 2 m/s as soon as the Frisbee passes her head how far would she have to run to catch it?

Homework Equations

x=x₀+V₀t+1/2at²

The Attempt at a Solution

So for part a, I figured if I made both x_frisbee and x_student equal to each other I could solve for t
x_frisbee=0+4t+(0.5)(-1.5)t²
x_student=0+0t+(0.5)(1)t²

0.5t²=4t+0.75t²
t=0s t=3.2s

And something similar for part b:
x_frisbee=0+4t+(0.5)(-1.5)t²
x_student=0+2t+(0.5)(0)t²

4t+(0.5)(-1.5)t²=2t
t=0s t=2.7s
x=(2m/s)(2.7s)=5.4m

I just wanted to know if I got these right. Thanks!

 

[Doc Al]

Looks perfect to me.

(Nitpick: Careful not to round off until the last step--especially if you're submitting your work to an online system. They can be picky.)

 

[jhae2.718]

Looks like you did everything correctly.

One caution: be careful with your signs; had you gone off of this equation, you would have gotten the result that the student will never reach the frisbee:

0.5t²=4t+0.75t²

What Doc Al says about rounding is good advice. The distance, if you carry figures through to the end, ends up at about 5.3333 m.

 

[dramallama]

Thank you for your help.

I'll make sure to hold the rounding off until the end and to check my signs next time.