Question

A body moves from rest with a uniform acceleration of $4\text{ m/s}^2$. The distance covered by it in the 5th second of its motion is:

• A. 16 m
• B. 18 m
• C. 20 m
• D. 22 m Correct Answer: (B) 18 m

Hint:

For questions involving motion starting from rest ($u = 0$), the ratio of distances covered in successive seconds follows the sequence of odd numbers ($1 : 3 : 5 : 7 : 9 : \dots$). Since the distance in the 1st second is $\frac{a}{2} \times 1 = 2\text{ m}$, the distance in the 5th second will simply be the 5th odd number ($9$) multiplied by $2\text{ m}$, which immediately yields $18\text{ m}$!

Answer 1

The Correct Option is B

Step 1: Use the full distance formula at two times.
The distance in the 5th second is just the gap between how far the body has gone in 5 seconds and how far it has gone in 4 seconds. So I will find both total distances and subtract. The body starts from rest, so $u=0$ and $a=4\text{ m/s}^2$.

Step 2: Distance in 5 seconds.
Using $s=ut+\tfrac{1}{2}at^2$ with $u=0$ gives \[ s_5 = \tfrac{1}{2}\times 4 \times 5^2 = 2\times 25 = 50\text{ m} \]

Step 3: Distance in 4 seconds.
Same formula with $t=4$: \[ s_4 = \tfrac{1}{2}\times 4 \times 4^2 = 2\times 16 = 32\text{ m} \]

Step 4: Subtract to get the 5th second distance.
The distance covered only during the 5th second is the difference: \[ s_5 - s_4 = 50 - 32 = 18\text{ m} \]
So the body moves 18 m during the 5th second, which is option (B).
\[ \boxed{18\text{ m}} \]