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
a. Vo=56km/1h = 56000m/3600s=15.56m/s.
d = Vo*t + 0.5a*t^2 = 22 m.
15.56*2.15 + 0.5a*2.15^2 = 22
33.44 + 2.31a = 22
2.31a = 22-33.44 = -11.44
a = -4.95 m/s^2.
b. V=Vo + at=15.56 - 4.95*2.15=4.92 m/s.
To solve this problem, we can use the equations of motion for uniformly accelerated motion.
(a) To find the magnitude of the car's acceleration before impact, we can use the equation:
\[d = v_i t + \frac{1}{2} a t^2\]
where:
- \(d\) is the distance traveled by the car (in meters),
- \(v_i\) is the initial velocity of the car (in meters per second),
- \(t\) is the time taken for the car to come to a stop (in seconds), and
- \(a\) is the acceleration (in meters per second squared).
In this case, the initial velocity (\(v_i\)) is 56 km/h, which needs to be converted to m/s:
\[v_i = 56 \, \text{km/h} \times \frac{1000 \, \text{m}}{1 \, \text{km}} \times \frac{1 \, \text{h}}{3600 \, \text{s}} = 15.56 \, \text{m/s}\]
The distance traveled (\(d\)) is given as 22.0 m, and the time (\(t\)) is given as 2.15 s.
Plugging in these values into the equation above, we have:
\[22.0 \, \text{m} = 15.56 \, \text{m/s} \times 2.15 \, \text{s} + \frac{1}{2} a \times (2.15 \, \text{s})^2\]
Simplifying the equation, we can solve for the acceleration (\(a\)):
\[a = \frac{22.0 \, \text{m} - 15.56 \, \text{m/s} \times 2.15 \, \text{s}}{0.5 \times (2.15 \, \text{s})^2}\]
Calculating this expression will give us the magnitude of the car's acceleration before impact.
(b) Once we have the acceleration, we can find the final velocity of the car using the equation:
\[v_f = v_i + a \times t\]
where:
- \(v_f\) is the final velocity of the car (in meters per second),
- \(v_i\) is the initial velocity of the car (in meters per second),
- \(a\) is the acceleration (in meters per second squared), and
- \(t\) is the time taken for the car to come to a stop (in seconds).
Plugging in the values we already know, we can calculate the final velocity (\(v_f\)).
Let's calculate these values step by step.
To find the magnitude of the car's acceleration before impact, we can use the equations of motion. We know the initial velocity (v0) of the car, the distance (d) it traveled before hitting the barrier, and the time (t) it took to hit the barrier.
(a) To find the magnitude of acceleration (a), we can use the equation:
d = v0t + 0.5at^2
Rearranging the equation to solve for acceleration, we have:
a = (d - v0t) / (0.5t^2)
Plugging in the given values, we have:
d = 22.0 m (distance)
v0 = 56 km/h (initial velocity)
t = 2.15 s (time)
First, we need to convert the initial velocity from km/h to m/s:
v0 = 56 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 15.56 m/s
Now, we can substitute the values into the equation:
a = (22.0 m - 15.56 m/s * 2.15 s) / (0.5 * (2.15 s)^2)
Calculating the equation, we find:
a ≈ -1.26 m/s^2
Therefore, the magnitude of the car's acceleration before impact is approximately 1.26 m/s^2.
(b) To find the speed of the car at impact, we can use the following equation:
v = v0 + at
Plugging in the given values:
v0 = 15.56 m/s (initial velocity)
a = -1.26 m/s^2 (acceleration)
t = 2.15 s (time)
Substituting the values into the equation, we have:
v = 15.56 m/s + (-1.26 m/s^2) * 2.15 s
Calculating the equation, we find:
v ≈ 12.30 m/s
Therefore, the car was traveling at approximately 12.30 m/s at impact.