This problem involves calculating the braking force and time required to stop a car under different conditions. We will apply the work-energy principle and basic kinematic equations to solve the problem. Let's break it down step by step.
Given Data
- Initial speed of the car: 50 mi/h (≈ 80 km/h),
- Weight of the car: 3000 lb (≈ 13,000 N),
- Stopping distance: 200 ft (≈ 61 m).
We will use the following assumptions and conversions:
- 1 lb ≈ 4.448 N,
- 1 mi/h ≈ 0.44704 m/s.
Part (a): Braking Force
The work-energy principle states that the work done by the braking force is equal to the change in kinetic energy of the car. The car is brought to a stop, so the final velocity is 0 m/s. The initial kinetic energy is:
K.E. = (1/2) m v^2
Since the car's weight is given as 3000 lb (≈13,000 N), we can calculate the car's mass using the equation $F = mg$, where $g$ is the acceleration due to gravity (approximately 9.8 m/s²):
m = \frac{13,000}{9.8} ≈ 1326.5 \, \text{kg}
Now, the initial speed of the car is 50 mi/h, which is approximately 22.35 m/s (using the conversion factor 1 mi/h ≈ 0.44704 m/s). The initial kinetic energy is:
K.E. = (1/2) \times 1326.5 \, \text{kg} \times (22.35 \, \text{m/s})^2 ≈ 343,318 \, \text{J}
The work done by the braking force is equal to the change in kinetic energy. Since the car comes to rest, the work done by the braking force $W$ is:
W = F_{\text{brake}} \times d = 343,318 \, \text{J}
We are given that the stopping distance is 200 ft (≈ 61 m), so:
F_{\text{brake}} \times 61 = 343,318
Solving for $F_{\text{brake}}$:
F_{\text{brake}} ≈ \frac{343,318}{61} ≈ 5,633 \, \text{N}
So, the braking force required is approximately 5,633 N.
Part (b): Time Required to Stop
Now, we can use the equation of motion to calculate the time required to stop. We know the car's initial velocity \( v_i = 22.35 \, \text{m/s} \), the final velocity \( v_f = 0 \, \text{m/s} \), and the braking force. The acceleration \( a \) can be calculated using Newton’s second law:
F = ma, so a = \frac{F_{\text{brake}}}{m}
Substituting the values:
a = \frac{5,633 \, \text{N}}{1326.5 \, \text{kg}} ≈ -4.25 \, \text{m/s}^2
The negative sign indicates deceleration (slowing down). Now, using the kinematic equation:
v_f = v_i + a t Substituting $v_f = 0$, $v_i = 22.35 \, \text{m/s}$, and $a = -4.25 \, \text{m/s}^2$:
0 = 22.35 + (-4.25) t Solving for $t$:
t ≈ \frac{22.35}{4.25} ≈ 5.26 \, \text{seconds}
So, the time required to stop is approximately 5.26 seconds.
Part (c): Distance and Time Required to Stop at 25 mi/h
Now, let's repeat the calculations for the car moving at 25 mi/h (≈ 40 km/h), which is approximately 11.18 m/s. First, we’ll calculate the stopping distance, assuming the same braking force.
The initial kinetic energy at 25 mi/h is:
K.E. = (1/2) \times 1326.5 \, \text{kg} \times (11.18 \, \text{m/s})^2 ≈ 66,594 \, \text{J}
Since the braking force is the same, the work done by the braking force is again equal to the change in kinetic energy:
W = F_{\text{brake}} \times d = 66,594 \, \text{J}
We already know $F_{\text{brake}} ≈ 5,633 \, \text{N}$, so:
d = \frac{66,594}{5,633} ≈ 11.8 \, \text{m}
Thus, the stopping distance is approximately 11.8 meters. Next, we’ll calculate the time required to stop. Using the same acceleration as before:
a = \frac{F_{\text{brake}}}{m} ≈ -4.25 \, \text{m/s}^2
Using the kinematic equation again:
v_f = v_i + a t Substituting $v_f = 0$, $v_i = 11.18 \, \text{m/s}$, and $a = -4.25 \, \text{m/s}^2$:
0 = 11.18 + (-4.25) t Solving for $t$:
t ≈ \frac{11.18}{4.25} ≈ 2.63 \, \text{seconds}
So, the time required to stop at 25 mi/h is approximately 2.63 seconds.
Summary of Results
- (a) Braking force: \( F_{\text{brake}} ≈ 5,633 \, \text{N} \)
- (b) Time required to stop at 50 mi/h: \( t ≈ 5.26 \, \text{seconds} \)
- (c) Stopping distance at 25 mi/h: \( d ≈ 11.8 \, \text{m} \)
- (d) Time required to stop at 25 mi/h: \( t ≈ 2.63 \, \text{seconds} \)
