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 2.03 m/s2. The magnitude of the car's velocity at the end of stage 2 is 4.87 times greater than it is at the end of stage 1. Find the magnitude of the acceleration in stage 2.
Vf1 = Vo + at,
Vf1 = 0 + 2.03t = 2.03t.
Vf2 = 2.03t + 4.87*2.03t = 11.92t,
Vf2 = at = 11.92t,
at = 11.92t,
Divide both sides by t:
a = 11.92m/s^2=acceleration in stage 2.
To find the magnitude of the acceleration in stage 2, we need to use the formulas of motion and the given information.
Let's assume the magnitude of the car's velocity at the end of stage 1 is V1. Since the magnitude of the car's velocity at the end of stage 2 is 4.87 times greater than V1, the magnitude of the car's velocity at the end of stage 2 is 4.87V1.
In stage 1, the magnitude of the car's acceleration is given as 2.03 m/s^2. Let's denote it as a1.
During stage 1, the car starts from rest (initial velocity, u = 0) and accelerates to V1. We can use the first equation of motion to find V1 in terms of a1 and the time taken in stage 1, t1:
V1 = u + a1 * t1
V1 = 0 + 2.03 * t1
V1 = 2.03t1
Now, during stage 2, the car starts from V1 and accelerates to 4.87V1. Let's denote the magnitude of the acceleration in stage 2 as a2. We can use the second equation of motion to find a2 in terms of V1, the final velocity at the end of stage 2, and the time taken in stage 2, t2:
V2 = V1 + a2 * t2
4.87V1 = V1 + a2 * t2
4.87 = 1 + (a2 * t2) / V1
From the information given, we know that each stage occupies the same amount of time. Therefore, t1 = t2 = t (let's assume these equal durations as t).
Substituting V1 = 2.03t into the equation above:
4.87 = 1 + (a2 * t) / (2.03t)
4.87 = 1 + (a2 / 2.03)
Now, we can solve for a2:
4.87 - 1 = a2 / 2.03
3.87 = a2 / 2.03
a2 = 3.87 * 2.03
a2 = 7.8603 m/s^2
Therefore, the magnitude of the acceleration in stage 2 is 7.8603 m/s^2.