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 3.19 m/s2. The magnitude of the car's velocity at the end of stage 2 is 2.90 times greater than it is at the end of stage 1. Calculate the magnitude of the acceleration in stage 2.
To calculate the magnitude of the acceleration in stage 2, we need to use the equations of motion. Let's break down the problem step by step.
Step 1: Find the velocity at the end of stage 1.
Given:
- Initial velocity (u) = 0 m/s (as the automobile starts from rest)
- Magnitude of acceleration (a1) in stage 1 = 3.19 m/s^2
The equation linking velocity, acceleration, and time is:
v = u + at
Since the initial velocity (u) is zero:
v1 = 0 + a1 * t1
Step 2: Find the velocity at the end of stage 2 in terms of stage 1.
Given:
- Magnitude of velocity (v2) at the end of stage 2 is 2.90 times greater than it is at the end of stage 1.
So, v2 = (2.90 * v1)
Step 3: Finding the magnitude of acceleration (a2) in stage 2.
Since both stages occupy the same amount of time, the time (t1) for stage 1 is equal to the time (t2) for stage 2.
Since we know v1, v2, and t2 are related, we can use the equation:
v2 = v1 + a2 * t2
Replacing v2 with 2.90 * v1:
2.90 * v1 = v1 + a2 * t2
Step 4: Find the value of a2.
Rearranging the equation:
a2 * t2 = 2.90 * v1 - v1
Simplifying:
(a2 * t2) = 1.90 * v1
Dividing both sides by t2:
a2 = (1.90 * v1) / t2
Therefore, the magnitude of the acceleration in stage 2 (a2) is given by (1.90 * v1) / t2.