Question

Starting from rest, a car accelerates with uniform acceleration of \( 4 \, \text{m/s}^2 \) for some time after which it comes to rest with uniform deceleration of \( 6 \, \text{m/s}^2 \). If the total time of travel is 5 s, then the total distance travelled by the car is:

• A. 25 m
• B. 30 m
• C. 60 m
• D. 125 m

Hint:

When calculating total distances with varying accelerations and decelerations, always consider breaking the problem into distinct phases and applying kinematic equations to each phase.

Answer 1

The Correct Option is B

- First, calculate the time for acceleration (\( t_1 \)) and deceleration (\( t_2 \)) phases. Given the total time \( t_1 + t_2 = 5 \, \text{s} \).
- The acceleration \( t_1 = \frac{v}{4} \) and deceleration \( t_2 = \frac{v}{6} \), solve for \( v \): \[ \frac{v}{4} + \frac{v}{6} = 5 \Rightarrow v = 12 \, \text{m/s} \]
- Calculate the distances: \[ s_1 = \frac{1}{2} \times 4 \times (3)^2 = 18 \, \text{m}, \quad s_2 = 12 \times 2 - \frac{1}{2} \times 6 \times (2)^2 = 12 \, \text{m} \]
- Total distance traveled is \( s = 18 \, \text{m} + 12 \, \text{m} = 30 \, \text{m} \).