A vehicle with an initial speed of 70km/h, accelerates at 6000 km/h^2 on a straight pavement. How long will the car drive to end up with a speed of 120 km/h? What is the distance that the car travels within the timeframe?
A vehicle with an initial speed of 70km/h, accelerates at 6000 km/h^2 on a straight pavement. How long will the car drive to end up with a speed of 120 km/h? What is the distance that the car travels within the timeframe?
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![**Physics Problem on Motion with Constant Acceleration**
**Problem Statement:**
A vehicle with an initial speed of 70 km/h accelerates at 6000 km/h^2 on a straight pavement. How long will the car drive to end up with a speed of 120 km/h? What is the distance that the car travels within the timeframe?
**Solution:**
To solve this problem, we will use the equations of motion with constant acceleration.
Given Data:
- Initial speed (u): 70 km/h
- Final speed (v): 120 km/h
- Acceleration (a): 6000 km/h^2
1. **Time Calculation:**
Using the first equation of motion:
\[ v = u + at \]
Rearrange to solve for time (t):
\[ t = \frac{v - u}{a} \]
Substitute the given values:
\[ t = \frac{120 \ \text{km/h} - 70 \ \text{km/h}}{6000 \ \text{km/h}^2} \]
\[ t = \frac{50 \ \text{km/h}}{6000 \ \text{km/h}^2} \]
\[ t \approx 0.00833 \ \text{h} \]
Convert hours to seconds:
\[ t \approx 0.00833 \ \text{h} \times 3600 \ \text{s/h} \]
\[ t \approx 30 \ \text{s} \]
2. **Distance Calculation:**
Using the second equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]
Substitute the given values:
\[ s = 70 \ \text{km/h} \times 0.00833 \ \text{h} + \frac{1}{2} \times 6000 \ \text{km/h}^2 \times (0.00833 \ \text{h})^2 \]
\[ s = 0.583 + \frac{1}{2} \times 6000 \times 0.0000694 \]
\[ s = 0.583 + 0.208 \]
\[ s \approx 0.791 \ \text{km} \]
**Summary:**
- The car will take approximately 30 seconds to reach a speed of 120 km/h.
- During this](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc2cb01cc-2ce0-42c9-8bdf-028787007c87%2F6a7970e3-12f9-40d2-b53b-709369bc5b7b%2F72fpzoj_processed.png&w=3840&q=75)
Transcribed Image Text:**Physics Problem on Motion with Constant Acceleration**
**Problem Statement:**
A vehicle with an initial speed of 70 km/h accelerates at 6000 km/h^2 on a straight pavement. How long will the car drive to end up with a speed of 120 km/h? What is the distance that the car travels within the timeframe?
**Solution:**
To solve this problem, we will use the equations of motion with constant acceleration.
Given Data:
- Initial speed (u): 70 km/h
- Final speed (v): 120 km/h
- Acceleration (a): 6000 km/h^2
1. **Time Calculation:**
Using the first equation of motion:
\[ v = u + at \]
Rearrange to solve for time (t):
\[ t = \frac{v - u}{a} \]
Substitute the given values:
\[ t = \frac{120 \ \text{km/h} - 70 \ \text{km/h}}{6000 \ \text{km/h}^2} \]
\[ t = \frac{50 \ \text{km/h}}{6000 \ \text{km/h}^2} \]
\[ t \approx 0.00833 \ \text{h} \]
Convert hours to seconds:
\[ t \approx 0.00833 \ \text{h} \times 3600 \ \text{s/h} \]
\[ t \approx 30 \ \text{s} \]
2. **Distance Calculation:**
Using the second equation of motion:
\[ s = ut + \frac{1}{2}at^2 \]
Substitute the given values:
\[ s = 70 \ \text{km/h} \times 0.00833 \ \text{h} + \frac{1}{2} \times 6000 \ \text{km/h}^2 \times (0.00833 \ \text{h})^2 \]
\[ s = 0.583 + \frac{1}{2} \times 6000 \times 0.0000694 \]
\[ s = 0.583 + 0.208 \]
\[ s \approx 0.791 \ \text{km} \]
**Summary:**
- The car will take approximately 30 seconds to reach a speed of 120 km/h.
- During this
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