. A 1200 kg car is moving at 95 km/h when the driver notices a deer down the road. He immediately moves his
foot toward the brake pedal, taking only 0.50 s before the car starts slowing down. The brakes cause a net force
of 2400 N [backwards] on the car for 2.0 s. The deer then jumps out of the way and the driver lifts his foot off
the brake pedal. How far does the car move in the 2.5 s starting from when the driver sees the deer?
Well, it seems like the car and the deer are about to play a little game of "dodge the brakes". Let's see how it goes.
First, we need to find out how much the car slows down during the 2.0 seconds when the brakes are applied. We can use Newton's second law, which states that force equals mass times acceleration.
So, the acceleration of the car is the net force divided by the mass: a = F/m. Plugging in the numbers, we get a = 2400 N / 1200 kg = 2 m/s².
Now, we can calculate how much the car slows down during those 2.0 seconds. We'll use the formula: Δv = a × t, where Δv is the change in velocity, a is the acceleration, and t is the time.
Δv = 2 m/s² × 2.0 s = 4 m/s.
Got it so far? Great! Now, let's move on to what happens after the deer jumps out of the way.
The driver lifts his foot off the brake pedal, which means there is no longer a net force acting on the car. So, the car will continue moving at the same velocity it had right after the brakes were applied.
During this 2.5-second interval, the car continues moving at a velocity of 4 m/s. And since we know that velocity equals distance divided by time, we can rearrange the formula to calculate the distance traveled.
Distance = velocity × time = 4 m/s × 2.5 s = 10 m.
So, the car moves a total distance of 10 meters starting from when the driver sees the deer. Hopefully, that was enough for the driver to avoid hitting the deer and give them both a fright. Stay safe on the roads and beware of deer with a penchant for acrobatics!
To find the distance the car moves in 2.5 seconds starting from when the driver sees the deer, we need to calculate the total distance covered during each phase: the initial acceleration, deceleration with the brakes, and final coasting.
1. Initial acceleration:
First, we need to calculate the initial velocity (v) of the car after the driver moves his foot toward the brake pedal in 0.50 seconds.
Using the formula: v = u + at, where u is the initial velocity, a is the acceleration, and t is the time,
Initial velocity = 95 km/h = 26.39 m/s (converting km/h to m/s)
Final velocity (assuming the car comes to a stop) = 0 m/s
Acceleration (a) = (final velocity - initial velocity) / time
Acceleration = (0 m/s - 26.39 m/s) / 0.50 s
Acceleration = -52.78 m/s² (negative sign indicates deceleration)
Next, we need to calculate the distance covered during this initial acceleration phase.
Using the formula: s = ut + (1/2)at², where s is the distance, u is the initial velocity, a is the acceleration, and t is the time,
Distance covered during acceleration = ut + (1/2)at²
Distance covered during acceleration = 26.39 m/s * 0.50 s + (1/2) * -52.78 m/s² * (0.50 s)²
Distance covered during acceleration = 13.2 m - 6.6 m
Distance covered during acceleration = 6.6 m
2. Deceleration with brakes:
During this phase, the car experiences a constant net force of 2400 N backwards for 2 seconds.
Using Newton's second law: F = ma, where F is the force, m is the mass, and a is the acceleration,
Acceleration = F / m
Acceleration = 2400 N / 1200 kg
Acceleration = 2 m/s²
Next, we need to calculate the distance covered during this deceleration phase.
Using the formula: s = ut + (1/2)at², where s is the distance, u is the initial velocity, a is the acceleration, and t is the time,
Distance covered during deceleration = ut + (1/2)at²
Distance covered during deceleration = 0 m/s * 2 s + (1/2) * -2 m/s² * (2 s)²
Distance covered during deceleration = 0 m - 4 m
Distance covered during deceleration = -4 m
3. Coasting:
After the deer jumps out of the way, the driver lifts his foot off the brake pedal. The car will continue to move at a constant velocity during this phase.
The car is already stopped, so we assume the final velocity is 0 m/s, and the acceleration is also 0 m/s².
Finally, we need to calculate the distance covered during this coasting phase.
Using the formula: s = ut + (1/2)at², where s is the distance, u is the initial velocity (0 m/s), a is the acceleration (0 m/s²), and t is the time,
Distance covered during coasting = ut + (1/2)at²
Distance covered during coasting = 0 m/s * 0.5 s + (1/2) * 0 m/s² * (0.5 s)²
Distance covered during coasting = 0 m + 0 m
Distance covered during coasting = 0 m
To find the total distance covered, we sum up the distances covered during each phase:
Total distance = Distance covered during acceleration + Distance covered during deceleration + Distance covered during coasting
Total distance = 6.6 m + (-4 m) + 0 m
Total distance = 2.6 m
Therefore, the car moves a distance of 2.6 meters in the 2.5 seconds starting from when the driver sees the deer.
To find the distance the car moves in 2.5 seconds, we need to break down the problem into different stages and calculate the distance for each stage.
First, let's calculate the initial velocity of the car when the driver starts applying the brakes:
Given:
Mass of the car (m) = 1200 kg
Initial velocity (u) = 95 km/h = 95 * 1000 m/3600 s = 26.39 m/s
Time taken to apply brakes (t1) = 0.50 s
Using the formula for velocity (v = u + at), where a is the acceleration and t is the time, we can rearrange the formula to solve for acceleration:
a = (v - u) / t
a = (0 - 26.39) / 0.50
a ≈ -52.78 m/s^2 (negative sign indicates deceleration)
Now, let's calculate the distance covered during the deceleration phase:
Using the formula for distance (s = ut + (1/2)at^2), where s is the distance, u is the initial velocity, a is the acceleration, and t is the time. Since the car is decelerating, the acceleration will be negative (-52.78 m/s^2).
s1 = ut1 + (1/2)at1^2
s1 = 26.39 * 0.50 + (1/2) * (-52.78) * (0.50)^2
s1 = 13.19 - 3.29
s1 ≈ 9.90 m
Next, let's calculate the distance covered while the brakes are applied:
Given:
Net force exerted by the brakes (F) = 2400 N
Time brakes are applied (t2) = 2.0 s
Using Newton's second law of motion (F = ma), we can calculate the acceleration:
a = F / m
a = 2400 / 1200
a = 2.0 m/s^2
Now, let's calculate the distance covered during this phase:
s2 = ut2 + (1/2)at2^2
s2 = 0 * 2.0 + (1/2) * 2.0 * (2.0)^2
s2 = 0 + 2 * 2
s2 = 4.0 m
Lastly, let's calculate the distance covered after the brakes are released:
Given:
Time after releasing brakes (t3) = 2.5 s
Since the car is not experiencing any external force after releasing the brakes, it will continue to move at a constant velocity.
s3 = ut3
s3 = 0 * 2.5
s3 = 0
Now, let's find the total distance covered:
Total distance (s_total) = s1 + s2 + s3
s_total = 9.90 + 4.0 + 0
s_total ≈ 13.90 m
Therefore, the car moves approximately 13.90 meters in the 2.5 seconds starting from when the driver sees the deer.