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whit19 · Aug 29, 2012, 12:03 PM

An automobile starts from rest and accelerates to a final velocity in two stages along a straight road. Each stage occupies the same amount of time. In stage 1, the magnitude of the car's acceleration is 3.2 m/s2. The magnitude of the car's velocity at the end of stage 2 is 2.0 times greater than it is at the end of stage 1. Find the magnitude of the acceleration in stage 2.

ebaines · Aug 29, 2012, 12:05 PM

Have you tried setting up the equations for this? Remember $\Delta v = a t$ .

whit19 · Aug 29, 2012, 12:35 PM

a= 3.2 m/s^2
and
t=2.0
a= 3.2(2.0)
is this what I should be doing?

ebaines · Aug 29, 2012, 01:01 PM

Not quite, The change in velocity in phase 1 is $\Delta v_1 = v_1-v_0 = a_1t$ . Since $v_0$ = 0, you have
$v_1 = a_1t = 3.2\frac m {s^2} t$

The change in velocity in phase 2 is $\Delta v_2 = v_2-v_1= a_2t$ and you know that $v_2 = 2v_1$ .

Can you take it from here?

whit19 · Aug 30, 2012, 12:59 PM

Could you give me a walk through of all the steps up to the solution I am still unsure how to go about to solve this question.