Question
When a car was travelling on a certain road at a speed of 20 ms-1,the shortest distance it could stop using the brakes was 20m.when travelling at a speed of 60ms-1 how many times the maximum deceleration should be increased to stop in the same shortest distance
1.×2 2.×3 3.×6 4.×9 5.×12
1.×2 2.×3 3.×6 4.×9 5.×12
Answers
GPT-4o mini
To determine how many times the maximum deceleration should be increased to stop a car traveling at 60 m/s in the same shortest distance as when traveling at 20 m/s, we can use the principles of physics related to motion.
1. **Equations of motion**: The stopping distance \( s \) can be related to the initial speed \( u \) and the deceleration \( a \) using the equation:
\[
v^2 = u^2 + 2as
\]
where \( v \) is the final speed (0 m/s when coming to a stop).
Rearranging gives:
\[
s = \frac{u^2}{2a}
\]
2. **Initial conditions**:
- For the first case (initial speed \( u_1 = 20 \, \text{m/s} \), stopping distance \( s = 20 \, \text{m} \)):
\[
20 = \frac{(20)^2}{2a_1}
\]
Rearranging gives:
\[
a_1 = \frac{(20)^2}{2 \times 20} = \frac{400}{40} = 10 \, \text{m/s}^2
\]
- For the second case (initial speed \( u_2 = 60 \, \text{m/s} \)):
We want to stop in the same distance \( s = 20 \, \text{m} \):
\[
20 = \frac{(60)^2}{2a_2}
\]
Rearranging gives:
\[
a_2 = \frac{(60)^2}{2 \times 20} = \frac{3600}{40} = 90 \, \text{m/s}^2
\]
3. **Comparing deceleration**:
Now we can find how many times greater \( a_2 \) is compared to \( a_1 \):
\[
\text{Factor} = \frac{a_2}{a_1} = \frac{90}{10} = 9
\]
Thus, the maximum deceleration should be increased by a factor of \( 9 \) to stop the car traveling at \( 60 \, \text{m/s} \) in the same distance of \( 20 \, \text{m} \).
The answer is **4. ×9**.
1. **Equations of motion**: The stopping distance \( s \) can be related to the initial speed \( u \) and the deceleration \( a \) using the equation:
\[
v^2 = u^2 + 2as
\]
where \( v \) is the final speed (0 m/s when coming to a stop).
Rearranging gives:
\[
s = \frac{u^2}{2a}
\]
2. **Initial conditions**:
- For the first case (initial speed \( u_1 = 20 \, \text{m/s} \), stopping distance \( s = 20 \, \text{m} \)):
\[
20 = \frac{(20)^2}{2a_1}
\]
Rearranging gives:
\[
a_1 = \frac{(20)^2}{2 \times 20} = \frac{400}{40} = 10 \, \text{m/s}^2
\]
- For the second case (initial speed \( u_2 = 60 \, \text{m/s} \)):
We want to stop in the same distance \( s = 20 \, \text{m} \):
\[
20 = \frac{(60)^2}{2a_2}
\]
Rearranging gives:
\[
a_2 = \frac{(60)^2}{2 \times 20} = \frac{3600}{40} = 90 \, \text{m/s}^2
\]
3. **Comparing deceleration**:
Now we can find how many times greater \( a_2 \) is compared to \( a_1 \):
\[
\text{Factor} = \frac{a_2}{a_1} = \frac{90}{10} = 9
\]
Thus, the maximum deceleration should be increased by a factor of \( 9 \) to stop the car traveling at \( 60 \, \text{m/s} \) in the same distance of \( 20 \, \text{m} \).
The answer is **4. ×9**.