Question

A car at rest is accelerated at \(2 ms^{-2}\) for 1 minute and then retarded at \(2 ms^{-2}\) for 1 minute to attain rest. The distance travelled by the car is

• A. 3600 m
• B. 1800 m
• C. 9600 m
• D. 4800 m
• E. 7200 m

Hint:

If acceleration and deceleration are equal in magnitude and time, the distances covered in each phase are equal.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The motion is in two phases: acceleration and retardation. The distance in each phase can be calculated using equations of motion.

Step 2: Detailed Explanation:
Given: \(u = 0\), \(a = 2 m/s^2\), \(t = 1 \text{ minute} = 60 s\). Phase 1 (Acceleration): \[ s_1 = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 2 \times (60)^2 = 1 \times 3600 = 3600 \text{ m} \] Final velocity after acceleration, \(v = u + at = 0 + 2 \times 60 = 120 m/s\). Phase 2 (Retardation): Initial velocity for this phase is \(u' = 120 m/s\). Final velocity \(v' = 0\). Retardation \(a' = -2 m/s^2\). Time \(t' = 60 s\). Distance during retardation: \[ s_2 = u't' + \frac{1}{2}a't'^2 = (120 \times 60) + \frac{1}{2} \times (-2) \times (60)^2 = 7200 - 3600 = 3600 \text{ m} \] Alternatively, \(v'^2 - u'^2 = 2a's_2 \Rightarrow 0 - (120)^2 = 2(-2)s_2 \Rightarrow -14400 = -4s_2 \Rightarrow s_2 = 3600 m\). Total distance: \(s = s_1 + s_2 = 3600 + 3600 = 7200 \text{ m}\).

Step 3: Final Answer:
The total distance travelled is 7200 m.