Question

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

• A. 60 m
• B. 25 m
• C. 36 m
• D. 66 m
• E. 45 m

Hint:

Always list out your known variables ($u$, $v$, $a$, $s$, $t$) before solving equations of motion . This helps in identifying the most suitable kinematic equation and avoids simple substitution errors. Check that all units are consistent (e.g., all in S.I. units) before calculating.

Answer 1

The Correct Option is D

Concept: The motion of an object under constant acceleration can be described using kinematic equations. The distance covered is a function of initial velocity, time, and the rate of acceleration.
• Second Equation of Motion: $s = ut + \frac{1}{2}at^2$
• $u$: Initial velocity
• $a$: Constant acceleration
• $t$: Time duration

Step 1: {Identify the given parameters from the problem statement.}
Given:
• Initial velocity ($u$) = 5 m/s
• Acceleration ($a$) = 2 m/s$^2$
• Time ($t$) = 6 s

Step 2: {Apply the distance formula for uniform acceleration.}
Using the formula: $$s = ut + \frac{1}{2}at^2$$

Step 3: {Substitute the known values into the equation.}
$$s = (5 \times 6) + \frac{1}{2} \times 2 \times (6)^2$$

Step 4: {Perform the final calculation to find distance.}
$$s = 30 + \frac{1}{2} \times 2 \times 36$$ $$s = 30 + 36$$ $$s = 66 \text{ m}$$