Find acceleration from distance and time, with a twist

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To find the acceleration of the drag racer who covered the last 40 meters in 0.20 seconds, the kinematic equations can be applied, considering constant acceleration. The hint provided indicates that the second driver's time is 0.05 seconds longer than the first, leading to a calculation of their acceleration using the same principles. By working backwards and applying the kinematic formulas, the distance ahead of the second driver at the finish line can also be determined. The problem emphasizes the importance of understanding one-dimensional motion and the careful application of kinematics. Overall, the discussion revolves around solving a straightforward physics problem with a focus on acceleration and time.
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Homework Statement


A drag racer starts from rest and covers the last 40 meters of a 400 meter race in 0.20 seconds.
a) What was his acceleration
b) He beat another driver by 0.05 seconds, what was the second drivers acceleration?
c) How far was the first driver ahead when he crossed the finish line?

Acceleration is constant for both drag racers.

Homework Equations



All kinematics equations can be used.

The Attempt at a Solution



I tried to create an equation using variables but this proved tricky. I am unsure of where to start and how to proceed. Also we were given a hint: t2=t1-0.20

__________t1____________
---------------------------
_________t2______/(40 m)/
 
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Conceptually this is a simple problem - it just require the careful application of some basic facts about one-dimensional motion and knowledge of the basic kinematics formulas. It's probably easiest to work backwards.

You'll need the following formulas:

$$Δx = v_{1} Δt + \frac{1}{2} a (Δt)^{2}$$
$$v_{2} = v_{1} + a Δt$$
 
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