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 driver's encounter with the deer had quite the impact. Let's calculate the distance traveled by the car during this event, shall we?
First, we need to determine the initial velocity of the car. Since it was moving at 95 km/h, we'll convert that to m/s: 95 km/h x (1000 m/1 km) x (1 h/3600 s) = 26.4 m/s. Got it? Great!
Now, let's find out how much the car slows down in the 2.0 s when the brakes are applied. We can use Newton's second law: F = ma, where F is the net force and a is the acceleration. Rearranging the equation to solve for acceleration gives us: a = F/m = 2400 N / 1200 kg = 2 m/s². Good so far?
To find the distance traveled during the braking period, we can use the equation: d = Vit + 0.5at². With an initial velocity of 26.4 m/s, a time of 2.0 s, and an acceleration of -2 m/s² (negative because it's slowing down), plugging in the values gives us: d = (26.4 m/s)(2.0 s) + 0.5(-2 m/s²)(2.0 s)² = 52.8 m - 4 m = 48.8 m. Are you following along?
Finally, we need to calculate the distance traveled after the brakes are released. Since the car is no longer accelerating or decelerating, the distance can be calculated using the equation: d = Vit, where the initial velocity is 0 m/s (since it's not moving), and the time is 0.5 s. Therefore, d = (0 m/s)(0.5 s) = 0 m. Easy-peasy!
Adding up the distances traveled during the braking period and after the brakes were released, we get: 48.8 m + 0 m = 48.8 m. So, the car moves a total distance of 48.8 meters during the 2.5-second period. Don't worry, the deer never had a chance to teach the driver any new dance moves!
To find how far the car moves in the 2.5 s starting from when the driver sees the deer, we need to calculate the distance covered during each stage of the motion: acceleration, deceleration, and constant velocity.
1. Acceleration phase:
We will first calculate the final velocity after the acceleration phase using the formula:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Given:
Initial velocity (u) = 95 km/h = (95 * 1000) / 3600 m/s ≈ 26.39 m/s
Acceleration (a) = Net force / mass
Mass (m) = 1200 kg
Net force (F) = 2400 N [backwards]
a = F / m = 2400 N / 1200 kg = 2 m/s² (taking direction into consideration)
Using the formula, we can find the final velocity (v):
v = u + at
v = 26.39 m/s + 2 m/s² * 0.50 s
v ≈ 27.39 m/s
Now, we need to calculate the distance covered during this acceleration phase using the formula:
s = ut + (1/2)at²
s = 26.39 m/s * 0.50 s + (1/2) * 2 m/s² * (0.50 s)²
s ≈ 13.20 m
2. Deceleration phase:
Next, we will calculate the distance covered during the deceleration phase using the formula:
s = vt - (1/2)at²
Here, the initial velocity (u) is the final velocity from the acceleration phase, and time (t) is given as 2.0 s.
Let's calculate the final velocity (v) during this deceleration phase:
v = u + at
v = 27.39 m/s + (-2 m/s²) * 2.0 s
v ≈ 23.39 m/s
Using the formula, we can find the distance covered during deceleration:
s = 27.39 m/s * 2.0 s - (1/2) * (-2 m/s²) * (2.0 s)²
s ≈ 44.78 m
3. Constant velocity phase:
Finally, during this phase, the velocity is constant at 23.39 m/s. The distance covered during constant velocity can be calculated using:
s = vt
s = 23.39 m/s * 0.5 s
s ≈ 11.70 m
Now we can calculate the total distance covered:
Total distance = distance during acceleration + distance during deceleration + distance during constant velocity
Total distance ≈ 13.20 m + 44.78 m + 11.70 m
Total distance ≈ 69.68 m
Therefore, the car moves approximately 69.68 meters in the 2.5 seconds starting from when the driver sees the deer.
To find the distance the car moves, we can use the kinematic equation:
\(d = v_i t + \frac{1}{2} a t^2\)
where:
- \(d\) is the distance traveled
- \(v_i\) is the initial velocity
- \(t\) is the time
- \(a\) is the acceleration
Now, let's break down the problem step by step to find the distance.
Step 1: Find the initial velocity
The initial velocity (\(v_i\)) is given as 95 km/h. However, we need to convert it to meters per second (m/s) to be consistent with the other units.
\(v_i\) = 95 km/h * (1000 m/km) / (3600 s/h) = 26.39 m/s
Step 2: Find the acceleration
To find the acceleration (\(a\)), we need to use Newton's second law:
\(F = ma\)
where:
- \(F\) is the net force applied to the car
- \(m\) is the mass of the car
The net force applied to the car is given as 2400 N (backwards). According to Newton's second law, this force equals mass times acceleration:
2400 N = (1200 kg) * a
Solving for \(a\):
\(a = \frac{2400 N}{1200 kg} = 2.0 m/s^2\)
Step 3: Calculate the distance
Now we can plug in the values of \(v_i\), \(t\), and \(a\) into the kinematic equation to find the distance traveled:
\(d = v_i t + \frac{1}{2} a t^2\)
\(d = (26.39 m/s) * (2.5 s) + \frac{1}{2} (2.0 m/s^2) * (2.5 s)^2\)
\(d = 65.98 m + 6.25 m\)
\(d = 72.23 m\)
Therefore, the car moves approximately 72.23 meters in the 2.5 seconds starting from when the driver sees the deer.