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:
This is a kinematics problem involving two phases of motion: constant acceleration followed by constant retardation (deceleration). We need to calculate the distance covered in each phase and add them to find the total distance.
Step 2: Key Formula or Approach:
We will use the equations of motion for constant acceleration:
\(v = u + at\)
\(s = ut + \frac{1}{2}at^2\)
\(v^2 = u^2 + 2as\)
where u = initial velocity, v = final velocity, a = acceleration, t = time, and s = distance.
Step 3: Detailed Explanation:
First, convert the time from minutes to seconds: t = 1 minute = 60 s.
Phase 1: Acceleration
Initial velocity, \(u_1 = 0\) (starts from rest)
Acceleration, \(a_1 = 2\) m/s\(^2\)
Time, \(t_1 = 60\) s
Let's find the distance travelled in this phase (\(s_1\)) and the final velocity (\(v_1\)). Using the second equation of motion for distance: \[ s_1 = u_1t_1 + \frac{1}{2}a_1t_1^2 \] \[ s_1 = (0)(60) + \frac{1}{2}(2)(60)^2 \] \[ s_1 = (1)(3600) = 3600 \text{ m} \] Now, let's find the velocity at the end of this phase, which will be the initial velocity for the next phase. Using the first equation of motion: \[ v_1 = u_1 + a_1t_1 \] \[ v_1 = 0 + (2)(60) = 120 \text{ m/s} \] Phase 2: Retardation
Initial velocity, \(u_2 = v_1 = 120\) m/s
Retardation, so acceleration is negative, \(a_2 = -2\) m/s\(^2\)
Time, \(t_2 = 60\) s
Final velocity, \(v_2 = 0\) (comes to rest). We can use this to check our calculation. \(v = u + at = 120 + (-2)(60) = 120 - 120 = 0\). This confirms the time is correct.
Let's find the distance travelled in this phase (\(s_2\)). Using the second equation of motion: \[ s_2 = u_2t_2 + \frac{1}{2}a_2t_2^2 \] \[ s_2 = (120)(60) + \frac{1}{2}(-2)(60)^2 \] \[ s_2 = 7200 - (1)(3600) = 3600 \text{ m} \] Total Distance The total distance travelled is the sum of the distances from both phases. \[ S_{total} = s_1 + s_2 = 3600 \text{ m} + 3600 \text{ m} = 7200 \text{ m} \] Step 4: Final Answer:
The total distance travelled by the car is 7200 m. This corresponds to option (E).