Question
A car is being driven along a road with a steady speed of 25 ms ^-1 when the driver suddenly notices that there is a fallen tree blocking the road 65 m ahead the driver immediately applies the Brake giving the car a constant retardation of 5ms^-2. How far in front of the car does the tree comes to rest. If the driver had not reached immediately and the brakes were applied one second later with what speed would the Car have hit the tree
Answers
GPT-4o mini
To solve the problem, we can break it into two parts: first, we will find out how far the car travels before it comes to a stop after the driver applies the brake immediately. Then, we'll analyze what happens if the brakes are applied one second later.
### Part 1: Immediate Braking
1. **Given Data**:
- Initial speed (\(u\)) = 25 m/s
- Retardation (\(a\)) = -5 m/s² (negative because it's deceleration)
- Distance to the tree = 65 m
2. **Using the equation of motion**:
\[
v^2 = u^2 + 2as
\]
where:
- \(v\) = final velocity (0 m/s when the car stops)
- \(s\) = distance traveled under deceleration
- Rearranging gives us:
\[
0 = (25)^2 + 2(-5)s
\]
\[
0 = 625 - 10s \\
10s = 625 \\
s = 62.5 \, \text{m}
\]
3. **Distance to the tree**:
- The car travels 62.5 m while it comes to rest.
- Since the distance to the tree is 65 m, the distance in front of the car where it comes to rest is:
\[
65 \, \text{m} - 62.5 \, \text{m} = 2.5 \, \text{m}
\]
### Part 2: Delayed Braking
- If the driver waits 1 second before braking, we first calculate how far the car travels in that second:
1. Distance traveled in one second at constant speed:
\[
\text{Distance} = \text{speed} \times \text{time} = 25 \, \text{m/s} \times 1 \, \text{s} = 25 \, \text{m}
\]
- New distance to the tree after 1 second:
\[
65 \, \text{m} - 25 \, \text{m} = 40 \, \text{m}
\]
2. **Now we find the speed of the car when the driver applies the brakes (after 1 second)**:
- After 1 second, the speed of the car:
\[
v = u + at = 25 \, \text{m/s} + 0 \cdot 1 \, \text{s} = 25 \, \text{m/s}
\]
3. **Applying the equation of motion again**:
\[
v^2 = u^2 + 2as
\]
where now the initial speed (\(u\)) is 25 m/s and \(s\) is now 40 m:
\[
v^2 = (25)^2 + 2(-5)(40)
\]
\[
v^2 = 625 - 400
\]
\[
v^2 = 225
\]
\[
v = \sqrt{225} = 15 \, \text{m/s}
\]
### Conclusion
1. When the driver applies the brakes immediately, the car comes to rest 2.5 m in front of the tree.
2. If the driver had waited 1 second before applying the brakes, the car would have hit the tree at a speed of 15 m/s.
### Part 1: Immediate Braking
1. **Given Data**:
- Initial speed (\(u\)) = 25 m/s
- Retardation (\(a\)) = -5 m/s² (negative because it's deceleration)
- Distance to the tree = 65 m
2. **Using the equation of motion**:
\[
v^2 = u^2 + 2as
\]
where:
- \(v\) = final velocity (0 m/s when the car stops)
- \(s\) = distance traveled under deceleration
- Rearranging gives us:
\[
0 = (25)^2 + 2(-5)s
\]
\[
0 = 625 - 10s \\
10s = 625 \\
s = 62.5 \, \text{m}
\]
3. **Distance to the tree**:
- The car travels 62.5 m while it comes to rest.
- Since the distance to the tree is 65 m, the distance in front of the car where it comes to rest is:
\[
65 \, \text{m} - 62.5 \, \text{m} = 2.5 \, \text{m}
\]
### Part 2: Delayed Braking
- If the driver waits 1 second before braking, we first calculate how far the car travels in that second:
1. Distance traveled in one second at constant speed:
\[
\text{Distance} = \text{speed} \times \text{time} = 25 \, \text{m/s} \times 1 \, \text{s} = 25 \, \text{m}
\]
- New distance to the tree after 1 second:
\[
65 \, \text{m} - 25 \, \text{m} = 40 \, \text{m}
\]
2. **Now we find the speed of the car when the driver applies the brakes (after 1 second)**:
- After 1 second, the speed of the car:
\[
v = u + at = 25 \, \text{m/s} + 0 \cdot 1 \, \text{s} = 25 \, \text{m/s}
\]
3. **Applying the equation of motion again**:
\[
v^2 = u^2 + 2as
\]
where now the initial speed (\(u\)) is 25 m/s and \(s\) is now 40 m:
\[
v^2 = (25)^2 + 2(-5)(40)
\]
\[
v^2 = 625 - 400
\]
\[
v^2 = 225
\]
\[
v = \sqrt{225} = 15 \, \text{m/s}
\]
### Conclusion
1. When the driver applies the brakes immediately, the car comes to rest 2.5 m in front of the tree.
2. If the driver had waited 1 second before applying the brakes, the car would have hit the tree at a speed of 15 m/s.