Kinematics

We are given the following information: - The car starts from rest, so initial velocity $u%20%3D%200$ .
- The final velocity $v%20%3D%20180%20%5C%2C%20%5Ctext%7Bkm%2Fh%7D%20%3D%20180%20%5Ctimes%20%5Cfrac%7B1000%7D%7B3600%7D%20%5C%2C%20%5Ctext%7Bm%2Fs%7D%20%3D%2050%20%5C%2C%20%5Ctext%7Bm%2Fs%7D$ .
- The time interval $t%20%3D%2010%20%5C%2C%20%5Ctext%7Bs%7D$ . We need to find the distance covered by the car in this time.
Using the equation of motion for uniformly accelerated motion: $s%20%3D%20ut%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20a%20t%5E2$
Since the initial velocity $u%20%3D%200$ , the equation simplifies to: $s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20a%20t%5E2$
To find the acceleration $a$ , we use the equation: $v%20%3D%20u%20%2B%20at$ Substitute the known values: $50%20%3D%200%20%2B%20a%20%5Ctimes%2010$ Solving for $a$ : $a%20%3D%20%5Cfrac%7B50%7D%7B10%7D%20%3D%205%20%5C%2C%20%5Ctext%7Bm%2Fs%7D%5E2$ Now, substitute $a%20%3D%205%20%5C%2C%20%5Ctext%7Bm%2Fs%7D%5E2$ and $t%20%3D%2010%20%5C%2C%20%5Ctext%7Bs%7D$ into the equation for distance: $s%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%205%20%5Ctimes%2010%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%205%20%5Ctimes%20100%20%3D%20250%20%5C%2C%20%5Ctext%7Bm%7D$
Thus, the distance covered by the car is 250 m. Therefore, the correct answer is: $%5Ctext%7B(B)%20%7D%20250%20%5C%2C%20%5Ctext%7Bm%7D$

High-Yield Trend

Chapter Questions
3 MCQs

Official Solution

Official Solution

Official Solution

Kinematics

High-Yield Trend

Chapter Questions 3 MCQs

Official Solution

Official Solution

Official Solution

Chapter Questions
3 MCQs