Question

A car starts from rest and accelerates uniformly to a speed of 180 km/h in 10 s. The distance covered by the car in this time interval is

• A. 500 m
• B. 250 m
• C. 100 m
• D. 200 m

Hint:

When solving for distance in uniformly accelerated motion, use the equation \( s = \frac{1}{2} a t^2 \), where \( a \) is the acceleration and \( t \) is the time. Ensure you convert all units to the standard SI units before calculating.

Answer 1

The Correct Option is B

We are given the following information: - The car starts from rest, so initial velocity \( u = 0 \). 
- The final velocity \( v = 180 \, \text{km/h} = 180 \times \frac{1000}{3600} \, \text{m/s} = 50 \, \text{m/s} \). 
- The time interval \( t = 10 \, \text{s} \). We need to find the distance covered by the car in this time. 
Using the equation of motion for uniformly accelerated motion: \[ s = ut + \frac{1}{2} a t^2 \] 
Since the initial velocity \( u = 0 \), the equation simplifies to: \[ s = \frac{1}{2} a t^2 \] 
To find the acceleration \( a \), we use the equation: \[ v = u + at \] Substitute the known values: \[ 50 = 0 + a \times 10 \] Solving for \( a \): \[ a = \frac{50}{10} = 5 \, \text{m/s}^2 \] Now, substitute \( a = 5 \, \text{m/s}^2 \) and \( t = 10 \, \text{s} \) into the equation for distance: \[ s = \frac{1}{2} \times 5 \times 10^2 = \frac{1}{2} \times 5 \times 100 = 250 \, \text{m} \] 
Thus, the distance covered by the car is 250 m. Therefore, the correct answer is: \[ \text{(B) } 250 \, \text{m} \]