A drag racer, starting from rest, speeds up for 402 m with an acceleration of +18.0 m/s2. A parachute then opens, slowing the car down with an acceleration of -6.30 m/s^2. How fast is the racer moving 2.70 x 10^2 m after the parachute opens?
find the velocity after 402m
vf^2=vi^2+2ad solve for Vf
Then, using that velocity for the initial velocity vi for the deacceleration time,
Vf^2=Vi^2+2ad where a=-6.30m/s^2 and d=2700m
To solve this problem, we need to break it down into two parts: the acceleration phase and the deceleration phase.
1. Acceleration Phase:
The initial velocity (u1) is 0 m/s.
The acceleration (a1) is +18.0 m/s^2.
The distance (s1) covered during this phase is 402 m.
To find the final velocity (v1) at the end of the acceleration phase, we can use the kinematic equation:
v1^2 = u1^2 + 2a1s1
Plugging in the given values, we have:
v1^2 = 0^2 + 2*(18.0)*(402)
v1^2 = 0 + 2*(18.0)*(402)
v1^2 = 0 + 14436
v1^2 = 14436
v1 ≈ 120.1 m/s
2. Deceleration Phase:
The initial velocity (u2) is v1 ≈ 120.1 m/s.
The acceleration (a2) is -6.30 m/s^2.
The distance (s2) covered during this phase is 2.70 × 10^2 m.
To find the final velocity (v2) at the end of the deceleration phase, we can use the same kinematic equation:
v2^2 = u2^2 + 2a2s2
Plugging in the given values, we have:
v2^2 = (120.1)^2 + 2*(-6.30)*(2.70 × 10^2)
v2^2 = 14424.01 - 3402.6
v2^2 = 11021.41
v2 ≈ 104.9 m/s
Therefore, the racer's speed 2.70 × 10^2 m after the parachute opens is approximately 104.9 m/s.
To solve this problem, we can use the equations of motion to find the car's final velocity.
First, we need to find the time it takes for the car to reach the point where the parachute opens. We can use the following equation:
vf = vi + at
Where:
vf = final velocity
vi = initial velocity (which is 0 m/s since the car starts from rest)
a = acceleration (+18.0 m/s^2 in this case)
t = time
Rearranging the equation to solve for time:
t = (vf - vi) / a
Since vi is 0 m/s, the equation becomes:
t = vf / a
Substituting in the given values:
t = (2.70 x 10^2 m) / (+18.0 m/s^2)
t = 15.0 s
Now that we know the time, we can find the velocity when the parachute opens. Again, we can use the equation:
vf = vi + at
Where:
vf = final velocity
vi = initial velocity (which is 0 m/s since the car starts from rest)
a = acceleration (-6.30 m/s^2 in this case)
t = time (which is 15.0 s)
vf = 0 + (-6.30 m/s^2) * (15.0 s)
vf = -94.5 m/s
Since the final velocity is negative, it means the car is moving in the opposite direction (slowing down).
Therefore, the speed of the racer 2.70 x 10^2 m after the parachute opens is 94.5 m/s (since speed is always positive).