Asked by Christina
A particle has a velocity of 18m/s at a certain time and 2.4s later is velocity is 30m/s in the opposite direction. What are the magnitude and direction of the particle's acceleration during the 2.4s interval?
I udes a=change in velocity /change in time and I got -20m/s^2 is this correct?
If we use
(v_f - v_o)/t = a where v_f=-30m/s, v_o=18m/s and t=2.4s then
a = (-30-18)/2.4 m/s^2
So yes, it looks correct to me.
I udes a=change in velocity /change in time and I got -20m/s^2 is this correct?
If we use
(v_f - v_o)/t = a where v_f=-30m/s, v_o=18m/s and t=2.4s then
a = (-30-18)/2.4 m/s^2
So yes, it looks correct to me.
Answers
Answered by
jin
-20m/s is the acceleration
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