A drag racer, starting from rest, speeds up for 402 m with an acceleration of +23.0 m/s2. A parachute then opens, slowing the car down with an acceleration of -5.10 m/s2. How fast is the racer moving 2.50 102 m after the parachute opens?
To solve this problem, we can divide it into three separate parts:
1. The initial acceleration phase
2. The distance covered during the acceleration phase
3. The deceleration phase after the parachute opens
Step 1: Calculate the final velocity after the initial acceleration phase.
Using the equation of motion, v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity (0 m/s), a is the acceleration (+23.0 m/s^2), and s is the distance covered during the acceleration phase (402 m), we can solve for v:
v^2 = 0^2 + 2(23.0)(402)
v^2 = 0 + 18492
v^2 = 18492
v = √18492
v = 136 m/s (rounded to 3 significant figures)
Step 2: Calculate the distance covered during the deceleration phase.
Using the equation of motion, v^2 = u^2 + 2as, where v is the final velocity after the initial acceleration phase (136 m/s), u is the initial velocity after the acceleration phase (136 m/s), a is the deceleration (-5.10 m/s^2), and s is the distance covered during the deceleration phase (unknown), we can solve for s:
(136)^2 = (136)^2 + 2(-5.10)s
18496 = 18496 - 10.2s
10.2s = 0
s = 0 m (since the deceleration phase stops the car)
Step 3: Calculate the final velocity after the deceleration phase.
Using the formula, v = u + at, where v is the final velocity after the deceleration phase (unknown), u is the initial velocity after the acceleration phase (136 m/s), a is the deceleration (-5.10 m/s^2), and t is the time taken during the deceleration phase (unknown), we can solve for v:
v = 136 + (-5.10)t
We know that the total distance covered from the start of the acceleration phase to the final position is 2.50 * 10^2 m (250 m). So, we can use the following equation to find t:
s = ut + (1/2)at^2
250 = 136t + (1/2)(-5.10)t^2
By solving this quadratic equation, we can find t and then substitute it back into the equation v = 136 + (-5.10)t to find the final velocity.
Unfortunately, without the specific coefficients of the quadratic equation, it is not possible to solve for the final velocity in this particular situation.
To find the final velocity of the drag racer 2.50 x 10^2 m after the parachute opens, we need to understand that the situation can be divided into two parts: the period before the parachute opens and the period after the parachute opens.
First, let's find the final velocity of the drag racer at the end of the period before the parachute opens.
Using the kinematic equation:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Since the drag racer is starting from rest, the initial velocity (u) is 0. The acceleration (a) is given as +23.0 m/s^2, and the displacement (s) is given as 402 m.
Plugging in these values, we have:
v^2 = 0^2 + 2 * 23.0 m/s^2 * 402 m
v^2 = 0 + 18504 m^2/s^2
v^2 = 18504 m^2/s^2
Taking the square root of both sides:
v = √(18504 m^2/s^2)
v ≈ 136 m/s (rounded to three significant figures)
Thus, at the moment the parachute opens, the drag racer is moving at approximately 136 m/s.
Now, let's find the final velocity of the drag racer 2.50 x 10^2 m after the parachute opens.
This time, the acceleration (a) is given as -5.10 m/s^2, and the initial velocity (u) is the final velocity just before the parachute opened, which is 136 m/s. The displacement (s) is given as 2.50 x 10^2 m.
Using the same kinematic equation:
v^2 = u^2 + 2as
Plugging in these values, we have:
v^2 = (136 m/s)^2 + 2 * (-5.10 m/s^2) * (2.50 x 10^2 m)
v^2 = 18496 m^2/s^2 + (-2550 m^2/s^2)
v^2 = 15946 m^2/s^2
Taking the square root of both sides:
v = √(15946 m^2/s^2)
v ≈ 126 m/s (rounded to three significant figures)
Therefore, the drag racer is moving at approximately 126 m/s 2.50 x 10^2 m after the parachute opens.