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‰ˆ -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‰ˆ -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‰ˆ 15.56 - 31.09
v β‰ˆ -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:

𝑑 = 𝑣0𝑑 + (1/2)π‘Žπ‘‘^2

where:
- 𝑑 is the distance traveled
- 𝑣0 is the initial velocity
- 𝑑 is the time elapsed
- π‘Ž is the acceleration

Let's go through the steps to find the answers to both parts:

(a) Finding the magnitude of acceleration (π‘Ž):
We are given:
- Initial velocity, 𝑣0 = 56 km/h = 56 * (1000/3600) m/s = 15.56 m/s
- Distance traveled, 𝑑 = 22.0 m
- Time elapsed, 𝑑 = 2.15 s

Using the kinematic equation, we can rearrange it to solve for acceleration π‘Ž:

𝑑 = 𝑣0𝑑 + (1/2)π‘Žπ‘‘^2

Substituting the given values:

22.0 = 15.56(2.15) + (1/2)π‘Ž(2.15)^2

Simplifying the equation:

22.0 = 33.484 + (2.15^2/2)π‘Ž

Rearranging the equation:

(2.15^2/2)π‘Ž = 22.0 - 33.484

(2.15^2/2)π‘Ž = -11.484

Now, let's solve for π‘Ž:

π‘Ž = (-11.484) * (2 / 2.15^2)

Calculating π‘Ž:

π‘Ž β‰ˆ -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:

𝑣 = 𝑣0 + π‘Žπ‘‘

We are given:
- Initial velocity, 𝑣0 = 56 km/h = 15.56 m/s
- Time elapsed, 𝑑 = 2.15 s
- Acceleration, π‘Ž = -3.70 m/sΒ²

Substituting the given values into the equation:

𝑣 = 15.56 + (-3.70)(2.15)

Calculating 𝑣:

𝑣 β‰ˆ 15.56 + (-7.955)

𝑣 β‰ˆ 7.61 m/s

Therefore, the car was traveling at approximately 7.61 m/s at the moment of impact.