A 1200kg 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 brak e pedal. How far does the car move in the 2.5 s starting from when the driver sees the deer?
To determine the distance the car moves in the 2.5 s starting from when the driver sees the deer, we can break it down into three parts:
1. The initial motion of the car before the brakes are applied.
2. The motion of the car when the brakes are applied.
3. The motion of the car after the brakes are released.
1. Initial Motion:
The car is moving at 95 km/h, which can be converted to m/s:
95 km/h × (1000 m/1 km) × (1 h/3600 s) = 26.39 m/s
To find the distance traveled during the initial motion, we use the formula:
distance = initial velocity × time
distance = 26.39 m/s × 0.50 s = 13.195 m
2. Motion with Brakes Applied:
The net force exerted by the brakes is 2400 N, which causes the car to decelerate.
Using Newton's second law of motion, F = ma, where F is the force, m is the mass, and a is the acceleration, we can calculate the acceleration of the car:
2400 N = (1200 kg) × a
a = 2400 N / 1200 kg
a = 2 m/s^2
To find the distance traveled during this time, we use the equation of motion:
distance = initial velocity × time + (1/2) × acceleration × time^2
distance = 26.39 m/s × 2.0 s + (1/2) × 2 m/s^2 × (2.0 s)^2
distance = 52.78 m + 2 m/s^2 × 4 s^2 / 2
distance = 52.78 m + 4 m
distance = 56.78 m
3. Motion after Brakes are Released:
After 2.0 s, the driver releases the brakes, and the car continues to move at a constant velocity.
The remaining time is 2.5 s - 0.50 s - 2.0 s = 0 s
Since the car moves at a constant velocity, the distance traveled during this time is:
distance = velocity × time
distance = 26.39 m/s × 0 s = 0 m
Therefore, the distance the car moves in the 2.5 s starting from when the driver sees the deer is:
13.195 m + 56.78 m + 0 m = 70.975 m (rounded to three decimal places).
To calculate the distance the car moves in 2.5 seconds, we need to break down the problem into two parts: the time it takes for the car to slow down and stop, and the time it takes for the car to accelerate back to its original speed.
First, let's find out how much the car decelerates within the initial 0.5 seconds.
Given:
Mass of the car (m) = 1200 kg
Initial velocity (u) = 95 km/h = 26.4 m/s (converted from km/h to m/s)
Time to start deceleration (t1) = 0.50 s
Net force during deceleration (F) = -2400 N (negative indicates a backward force)
Acceleration during deceleration (a) = F/m
Using Newton's second law of motion, we can calculate the acceleration during the initial deceleration phase:
a = F/m
a = -2400 N / 1200 kg
Calculating the acceleration:
a = -2 m/s²
Now, we can calculate the distance covered during the initial deceleration phase using the following kinematic equation:
s = ut + (1/2)at²
Given:
Initial velocity (u) = 26.4 m/s
Time to start deceleration (t1) = 0.50 s
Acceleration during deceleration (a) = -2 m/s²
Calculating the distance covered during the initial deceleration:
s = ut1 + (1/2)at1²
Next, we need to calculate the distance covered during the constant deceleration phase, which lasts for 2.0 seconds.
Given:
Acceleration during deceleration (a) = -2 m/s²
Time during constant deceleration (t2) = 2.0 s
Using the kinematic equation, we can calculate the distance covered during the constant deceleration:
s = ut2 + (1/2)at2²
Given:
Initial velocity during constant deceleration = 0 m/s (the car comes to a stop)
Next, we need to calculate the distance covered during the acceleration phase, which occurs when the driver lifts his foot off the brake pedal. The car will accelerate back to its original speed.
Given:
Final velocity during acceleration (v) = 26.4 m/s (the initial velocity)
We can calculate the distance during the acceleration phase using the kinematic equation:
s = vt + (1/2)at²
Given:
Time during acceleration (t3) = 0.5 seconds (since the total time is 2.5 seconds and the first 2 seconds were used for deceleration)
By summing these distances, we can find the total distance covered by the car:
Total distance = s1 + s2 + s3
Therefore, to calculate the distance the car moves in 2.5 seconds, we need to know the values of u, F, m, t1, t2, and t3.
To calculate the distance the car moves in the 2.5 seconds starting from when the driver sees the deer, we need to break down the motion into different stages.
1. Calculate the initial velocity of the car:
The car is moving at 95 km/h, so we need to convert it to m/s.
95 km/h = 95,000 m/3600 s ≈ 26.39 m/s
2. Calculate the elapsed time during the first stage (before the brakes are applied):
The driver takes 0.50 s to move his foot towards the brake pedal.
3. Calculate the time taken to slow down the car:
The brakes cause a net force of 2400 N on the car for 2.0 s.
4. Calculate the acceleration of the car:
Using Newton's second law, F = ma, we can rearrange the equation to find the acceleration, a = F/m.
a = 2400 N / 1200 kg = 2 m/s²
5. Calculate the deceleration distance:
Using the kinematic equation, v² = u² + 2as, where v is the final velocity (0 m/s), u is the initial velocity (26.39 m/s), a is the acceleration (-2 m/s² as it is in the opposite direction), and s is the distance to be calculated.
0² = 26.39² + 2(-2)s
0 = 696.48 - 4s
4s = 696.48
s = 696.48 / 4
s ≈ 174.12 m
6. Calculate the time taken for the deceleration:
Since the brakes are applied for 2.0 seconds, this is the time taken for deceleration.
7. Calculate the distance covered during deceleration:
Using the equation s = ut + (1/2)at², where u is the initial velocity (26.39 m/s), t is the time taken for deceleration (2.0 s), and a is the acceleration (-2 m/s²), we can calculate the distance covered during deceleration.
s = 26.39 * 2 + (1/2) * (-2) * (2^2)
s = 52.78 - 2
s = 50.78 m
8. Calculate the distance covered during the final 0.5 seconds:
The car slows down for 2.0 seconds, so it remains at rest for the remaining 0.5 seconds.
As it is at rest, the distance covered during these 0.5 seconds is 0.
9. Calculate the total distance covered:
The total distance covered is the sum of the distances covered during the deceleration and the final rest period.
Total distance = 174.12 m + 50.78 m + 0 m
Total distance ≈ 224.90 m
Therefore, the car moves approximately 224.90 m in the 2.5 seconds starting from when the driver sees the deer.