A car traveling 56 km/h is 22.0 m from a barrier when the driver slams on the brakes. The car hits the barrier 2.15 s later.
(a) Assuming constant acceleration, what was the magnitude of the car's acceleration before impact?
_______ m/s2
(b) How fast was the car traveling at impact?
________ m/s
To solve this problem, we can use the kinematic equations of motion. In particular, we can use the equation:
v = u + at
where:
v is the final velocity,
u is the initial velocity,
a is the acceleration, and
t is the time.
Let's break down the problem step by step:
(a) To find the magnitude of the car's acceleration before impact, we need to use the equation of motion for displacement:
s = ut + (1/2)at^2
Given:
Initial velocity, u = 56 km/h = 56 * (1000/3600) m/s = 15.56 m/s
Final velocity, v = 0 m/s (since the car comes to a stop)
Time, t = 2.15 s
Initial displacement, s = 22.0 m
We can rearrange the equation to solve for acceleration, a:
s = ut + (1/2)at^2
0 = 15.56 * 2.15 + (1/2)a * (2.15)^2
0 = 33.47 + 2.32a
Solving for a, we have:
2.32a = -33.47
a = -33.47 / 2.32
a β -14.47 m/s^2 (since acceleration is negative)
Therefore, the magnitude of the car's acceleration before impact is approximately 14.47 m/s^2.
(b) To find the speed of the car at impact, we can use the equation we derived earlier:
v = u + at
Given:
Initial velocity, u = 15.56 m/s (same as before)
Acceleration, a β -14.47 m/s^2 (same as before)
Time, t = 2.15 s
Plugging in these values into the equation, we get:
v = 15.56 + (-14.47) * 2.15
v β 15.56 - 31.09
v β -15.53 m/s (negative sign indicates the car is moving in the opposite direction)
Therefore, the speed of the car at impact is approximately 15.53 m/s.
To find the answers to both parts of the question, we can use the following kinematic equation:
π = π£0π‘ + (1/2)ππ‘^2
where:
- π is the distance traveled
- π£0 is the initial velocity
- π‘ is the time elapsed
- π is the acceleration
Let's go through the steps to find the answers to both parts:
(a) Finding the magnitude of acceleration (π):
We are given:
- Initial velocity, π£0 = 56 km/h = 56 * (1000/3600) m/s = 15.56 m/s
- Distance traveled, π = 22.0 m
- Time elapsed, π‘ = 2.15 s
Using the kinematic equation, we can rearrange it to solve for acceleration π:
π = π£0π‘ + (1/2)ππ‘^2
Substituting the given values:
22.0 = 15.56(2.15) + (1/2)π(2.15)^2
Simplifying the equation:
22.0 = 33.484 + (2.15^2/2)π
Rearranging the equation:
(2.15^2/2)π = 22.0 - 33.484
(2.15^2/2)π = -11.484
Now, let's solve for π:
π = (-11.484) * (2 / 2.15^2)
Calculating π:
π β -3.70 m/sΒ²
Therefore, the magnitude of the car's acceleration before impact is approximately 3.70 m/sΒ² (rounded to two decimal places).
(b) Finding the speed of the car at impact:
We can find the final speed of the car using the following formula:
π£ = π£0 + ππ‘
We are given:
- Initial velocity, π£0 = 56 km/h = 15.56 m/s
- Time elapsed, π‘ = 2.15 s
- Acceleration, π = -3.70 m/sΒ²
Substituting the given values into the equation:
π£ = 15.56 + (-3.70)(2.15)
Calculating π£:
π£ β 15.56 + (-7.955)
π£ β 7.61 m/s
Therefore, the car was traveling at approximately 7.61 m/s at the moment of impact.