A drag racer, starting from rest, speeds up for 402 m with an acceleration of +24.0 m/s2. A parachute then opens, slowing the car down with an acceleration of -5.40 m/s2. How fast is the racer moving 3.75 102 m after the parachute opens?
Please post your attempt at solution.
x=1/2(vo+v)t
vo1 +1/2at2
and vo2 +t i just don't know what numbers to put where
To find the final velocity of the racer, we can break the problem down into two parts: the initial acceleration phase and the deceleration phase.
1. Initial acceleration phase:
Given:
Initial velocity, u = 0 m/s
Acceleration, a1 = +24.0 m/s^2
Distance, d1 = 402 m
Using the kinematic equation:
v^2 = u^2 + 2ad
We can solve this equation to find the final velocity (v1) at the end of the acceleration phase:
v1^2 = 0^2 + 2 * 24.0 * 402
v1^2 = 0 + 19344
v1^2 = 19344
v1 = √19344
v1 ≈ 139 m/s
2. Deceleration phase:
Given:
Initial velocity, u2 = 139 m/s (the final velocity of the first phase)
Acceleration, a2 = -5.40 m/s^2
Distance, d2 = 3.75 * 10^2 m
Again, using the kinematic equation:
v^2 = u^2 + 2ad
We can solve this equation to find the final velocity (v2) at the end of the deceleration phase:
v2^2 = 139^2 + 2 * (-5.40) * (3.75 * 10^2)
v2^2 = 19321 - 4104
v2^2 = 15217
v2 = √15217
v2 ≈ 123.4 m/s
Therefore, the racer is moving at approximately 123.4 m/s 3.75 * 10^2 m after the parachute opens.