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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


[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? :)