**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^21. **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

Initial speed u = 70 km/h Acceleration a = 6000 km/h2 Final speed v = 120 km/h

Answered: A vehicle with an initial speed of…

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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?