Question:medium

An object having a velocity of 5 m/s is accelerated at the rate 2 m/s\(^2\) for 6s. Find the distance travelled during the period of acceleration.

Show Hint

The formula \( S = ut + \frac{1}{2} a t^2 \) is crucial for calculating the distance travelled under uniform acceleration. Remember to substitute the correct values for initial velocity, acceleration, and time.
Updated On: Apr 19, 2026
  • 60 m
  • 25 m
  • 36 m
  • 66 m
  • 45 m
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
This problem involves an object undergoing uniform linear acceleration.
We need to calculate the total displacement given the initial conditions and the acceleration period.
Step 2: Key Formula or Approach:
The distance travelled can be found using the second kinematic equation of motion.
The formula is \( s = ut + \frac{1}{2}at^{2} \), where \(u\) is initial velocity, \(a\) is acceleration, and \(t\) is time.
Step 3: Detailed Explanation:
The given values from the problem are:
Initial velocity, \( u = 5 \text{ m/s} \).
Acceleration, \( a = 2 \text{ m/s}^{2} \).
Time interval, \( t = 6 \text{ s} \).
Now, we substitute these parameters into the formula:
\[ s = (5)(6) + \frac{1}{2}(2)(6)^{2} \] \[ s = 30 + (1)(36) \] \[ s = 30 + 36 = 66 \text{ m} \] Step 4: Final Answer:
The distance travelled by the object during this period is 66 m.
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