Question
A car is moving at a velocity of 90km/h on a horizontal road. The brakes are applied and the car comes to rest after a distance of 60m. The mass of the car is 900Kg. Calculate:
1) the deceleration
2) the braking force
3) the time take to bring the car to rest
1) the deceleration
2) the braking force
3) the time take to bring the car to rest
Answers
GPT-4o mini
To solve the problem, we need to first convert the initial velocity from km/h to m/s, as the other measurements (distance, mass) are in the metric system.
1. **Convert velocity**:
\[
\text{Initial velocity } (u) = 90 \text{ km/h} = \frac{90 \times 1000}{3600} \text{ m/s} = 25 \text{ m/s}
\]
2. **Calculate the deceleration**:
We can use the kinematic equation:
\[
v^2 = u^2 + 2as
\]
where
- \( v \) is the final velocity (0 m/s, since the car comes to rest),
- \( u \) is the initial velocity (25 m/s),
- \( a \) is the acceleration (deceleration in this case, so it will be negative),
- \( s \) is the distance (60 m).
Plugging in the values:
\[
0 = (25)^2 + 2a(60)
\]
Simplifying this gives:
\[
0 = 625 + 120a
\]
Solving for \( a \):
\[
120a = -625 \implies a = -\frac{625}{120} \approx -5.21 \text{ m/s}^2
\]
Thus, the **deceleration** is approximately **5.21 m/s²**.
3. **Calculate the braking force**:
We can use Newton's second law:
\[
F = ma
\]
where
- \( m \) is the mass of the car (900 kg),
- \( a \) is the deceleration calculated earlier.
Plugging in the values:
\[
F = 900 \times (-5.21) \approx -4689 \text{ N}
\]
The negative sign indicates that the force is acting in the opposite direction of the motion. The **braking force** is approximately **4689 N**.
4. **Calculate the time taken to bring the car to rest**:
We can use the kinematic equation:
\[
v = u + at
\]
Solving for \( t \):
\[
0 = 25 + (-5.21)t
\]
This simplifies to:
\[
5.21t = 25 \implies t = \frac{25}{5.21} \approx 4.8 \text{ s}
\]
Thus, the **time taken** to bring the car to rest is approximately **4.8 seconds**.
### Summary:
1. Deceleration: \( \approx 5.21 \text{ m/s}^2 \)
2. Braking Force: \( \approx 4689 \text{ N} \)
3. Time to bring to rest: \( \approx 4.8 \text{ s} \)
1. **Convert velocity**:
\[
\text{Initial velocity } (u) = 90 \text{ km/h} = \frac{90 \times 1000}{3600} \text{ m/s} = 25 \text{ m/s}
\]
2. **Calculate the deceleration**:
We can use the kinematic equation:
\[
v^2 = u^2 + 2as
\]
where
- \( v \) is the final velocity (0 m/s, since the car comes to rest),
- \( u \) is the initial velocity (25 m/s),
- \( a \) is the acceleration (deceleration in this case, so it will be negative),
- \( s \) is the distance (60 m).
Plugging in the values:
\[
0 = (25)^2 + 2a(60)
\]
Simplifying this gives:
\[
0 = 625 + 120a
\]
Solving for \( a \):
\[
120a = -625 \implies a = -\frac{625}{120} \approx -5.21 \text{ m/s}^2
\]
Thus, the **deceleration** is approximately **5.21 m/s²**.
3. **Calculate the braking force**:
We can use Newton's second law:
\[
F = ma
\]
where
- \( m \) is the mass of the car (900 kg),
- \( a \) is the deceleration calculated earlier.
Plugging in the values:
\[
F = 900 \times (-5.21) \approx -4689 \text{ N}
\]
The negative sign indicates that the force is acting in the opposite direction of the motion. The **braking force** is approximately **4689 N**.
4. **Calculate the time taken to bring the car to rest**:
We can use the kinematic equation:
\[
v = u + at
\]
Solving for \( t \):
\[
0 = 25 + (-5.21)t
\]
This simplifies to:
\[
5.21t = 25 \implies t = \frac{25}{5.21} \approx 4.8 \text{ s}
\]
Thus, the **time taken** to bring the car to rest is approximately **4.8 seconds**.
### Summary:
1. Deceleration: \( \approx 5.21 \text{ m/s}^2 \)
2. Braking Force: \( \approx 4689 \text{ N} \)
3. Time to bring to rest: \( \approx 4.8 \text{ s} \)