Calculate the force needed to bring a 1070–kg car to rest from a speed of 89.0 km/h in a distance of 110 m (a fairly typical distance for a non-panic stop).
-2.97×103 N correct answer
Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.34 m. Calculate the force exerted on the car and compare it with the force found in part (a).
Force = -1.40×105 N
Ratio = 47.0
correct answer
what are the steps to get these answers?
force = mass * acceleration
now the kinematics
first units
89 km/h * 1000 meters/km * 1 h/3600 seconds
= 24.7 meters/second
average speed during stop = 24.7/2 = 12.36
so how long did it take?
t = 110 meters/12.36 m/s
= 8.9 seconds to stop
acceleration assumed constant so it is
a = -24.7 m/s / 8.9 s
a = -2.78 m/s^2 (about 1/3 acceleration of gravity)
F = -1070*2.78 = -2975 N
so we agree
Now do all that for 2.34 meters (OUCH!!)
t = 2.34 meters/12.36 m/s
= .1893 seconds
a = -24.7/.1893 = - 130 m/s^2
( about 13 times gravity !!! )
F = -1070 * 130 = -139,599 N
139599/2975 = 46.9
To calculate the force needed to bring the car to rest in both scenarios, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a):
F = m * a
In this case, we need to find the force (F) to bring the car to rest. The mass of the car (m) is given as 1070 kg.
Step 1: Convert the speed from km/h to m/s
In the first scenario, the car is traveling at a speed of 89.0 km/h. To calculate the force, we need to convert this speed to meters per second (m/s). The conversion formula is:
speed in m/s = speed in km/h * (1 km/3.6 m)
speed in m/s = 89.0 km/h * (1 km/3.6 m) = 24.72 m/s (rounded to two decimal places)
Step 2: Calculate the acceleration (a)
To find the acceleration, we need to use the formula:
acceleration (a) = change in velocity / time taken
In the first scenario, the car is coming to rest, so the change in velocity is the initial velocity (given by the speed we just calculated) minus the final velocity (which is 0 m/s since the car stops completely). The time taken is not given, but we can calculate it using the distance and the initial speed.
distance = average speed * time taken
time taken = distance / average speed
Using the given distance of 110 m and the initial speed of 24.72 m/s, we can calculate the time taken:
time taken = 110 m / 24.72 m/s ≈ 4.45 s (rounded to two decimal places)
Now we can calculate the acceleration:
acceleration = (0 m/s - 24.72 m/s) / 4.45 s ≈ -5.56 m/s² (rounded to two decimal places)
Step 3: Calculate the force (F)
Finally, we can calculate the force using the formula:
F = m * a
F = 1070 kg * -5.56 m/s² ≈ -5,952.20 N ≈ -2.97 × 10³ N
So, the force needed to bring the car to rest in the first scenario is approximately -2.97 × 10³ N.
Now let's move on to the second scenario:
Step 1: Calculate the acceleration (a)
In the second scenario, the car stops in a distance of 2.34 m. Since we don't have the time taken, we cannot directly calculate the acceleration using the formula mentioned above. However, we can use a different equation that relates distance, initial velocity, and acceleration:
velocity² = initial velocity² + 2 * acceleration * distance
Since the final velocity is 0 m/s (the car comes to a stop), the equation becomes:
0² = v² + 2 * a * 2.34 m
0 = v² + 4.68 a
The initial velocity (v) is the same as before, 24.72 m/s. Plugging in the values, we get:
0 = (24.72 m/s)² + 4.68 a
Solving for the acceleration (a), we find:
a = - (24.72 m/s)² / 4.68 ≈ -130.00 m/s² (rounded to two decimal places)
Step 2: Calculate the force (F)
Using the same formula as before:
F = m * a
F = 1070 kg * -130.00 m/s² ≈ -139,100 N ≈ -1.40 × 10⁵ N
So, the force exerted on the car when hitting the concrete abutment is approximately -1.40 × 10⁵ N.
Step 3: Calculate the ratio
To compare the forces found in part (a) and the force exerted on the car in this scenario, we can calculate the ratio:
Ratio = force in part (a) / force in this scenario
Ratio = (-2.97 × 10³ N) / (-1.40 × 10⁵ N) ≈ 0.0211 ≈ 47.0 (rounded to two decimal places)
Therefore, the force exerted on the car when hitting the concrete abutment is approximately 47 times greater than the force needed to bring the car to rest in the first scenario.