How does it become s^-3?

Nat200 · Apr 29, 2009, 10:33 AM

A particle's speed at time t is given by v= kt^2, where k is some constant. What are the dimensions of k? (L= length , T= time)

A. LT
B. LT^-1
C. LT^-2
D. LT^ -3

^ meaning raised to that power. Why is the answer D?
[k] = [v] / [t]^2 = (m/s) / s^2 I got to this point but my question is how does it become m * s^-3?? How does the s become ^ -3?

Thank you!

Unknown008 · Apr 29, 2009, 10:45 AM

According to me, the answer is correct, LT^-3

Originally Posted by Nat200

[k] = [v] / [t]^2 = (m/s) / s^2 I got to this point but my question is how does it become m * s^-3 ??? how does the s become ^ -3??

You did a slight mistake which revealed huge, lol.

Ok, enough, let's be serious...

[k] = [v] / [t]^2

$[k] = \frac{\frac{m}{s}} {s^2}$

$[k] = \frac{m}{s} /{s^2}$

$[k] = \frac{m}{s} \times\frac{1}{s^2}$

See it now? :)

Add your answer here.