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 -6.10 m/s2. How fast is the racer moving 2.55 102 m after the parachute opens?
Do you mean 255 m after the parachute opens?
The velocity when the chute opens is
V1 = sqrt(2aX) = 138.9 m/s
The velocity V2 after traveling a distance X' = 255 m is calculable with V2^2 - V1^2 = 2*a*X
where a = -6.1 m/s^2 and X = 255 m
To solve this problem, we can use the kinematic equation:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration
s = displacement
First, let's calculate the velocity after the acceleration phase. The racer starts from rest (u = 0) and accelerates for 402 m with an acceleration of +24.0 m/s^2.
v^2 = 0^2 + 2 * 24.0 * 402
v^2 = 0 + 19344
v = sqrt(19344) ≈ 139 m/s
So, after the acceleration phase, the racer is moving at a speed of approximately 139 m/s.
Next, we need to find the velocity after the deceleration phase. The parachute opens and slows the car down with an acceleration of -6.10 m/s^2. The displacement in this phase is given as 2.55 * 10^2 m.
Using the same kinematic equation:
v^2 = u^2 + 2as
Since we are seeking the final velocity (v), we can consider the initial velocity (u) as 139 m/s (the velocity after the acceleration phase). The acceleration in this phase is -6.10 m/s^2, and the displacement (s) is 2.55 * 10^2 m.
v^2 = 139^2 + 2 * (-6.10) * (2.55 * 10^2)
v^2 = 19321 - 3134.1
v = sqrt(19321 - 3134.1) ≈ 137 m/s
Therefore, the racer is moving at a speed of approximately 137 m/s, 2.55 * 10^2 m after the parachute opens.