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 3.3 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.
To find the magnitude of the acceleration in stage 2, let's break down the problem and solve it step by step.
Let's assume the magnitude of the car's velocity at the end of stage 1 is V. According to the problem, 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. Therefore, the magnitude of the car's velocity at the end of stage 2 would be 2V.
Now, we know that each stage occupies the same amount of time. Let's call this time t.
In stage 1, the magnitude of the car's acceleration is given as 3.3 m/s². Using the equation of motion:
V = u + at
where V is the final velocity, u is the initial velocity (which is 0 since the car starts from rest), a is the acceleration, and t is the time.
For stage 1:
V1 = 0 + (3.3)(t)
V1 = 3.3t
For stage 2:
V2 = V1 + (a2)(t)
V2 = (3.3t) + (a2)(t)
V2 = t(3.3 + a2)
Given that the magnitude of the car's velocity at the end of stage 2 (V2) is 2V, we can write:
2V = t(3.3 + a2)
Substituting V = 3.3t into the equation, we get:
2(3.3t) = t(3.3 + a2)
Simplifying the equation:
6.6t = 3.3t + a2t
3.3t = a2t
a2 = 3.3
Therefore, the magnitude of the acceleration in stage 2 (a2) is 3.3 m/s².