Question

A particle starts moving with acceleration 2m/sec². Distance travelled by it in 5th half second is
(1)1.25
(2)2.25
(3)6.25
(4)30.25
All are in meter

Answer 1

Answer on Question#54924 – Physics – Mechanics | Kinematics | Dynamics

(2) 2.25

Question

A particle starts moving with acceleration $2\frac{m}{s^2}$. Distance travelled by it in 5th half second is

(1) 1.25

(2) 2.25

(3) 6.25

(4) 30.25

All are in meter

Solution

Use the formula of distance travelled during first $t$ seconds:

$$L = \frac{a t^2}{2} + V_0 t,$$

where $L$ – distance $(m)$, $a$ – acceleration $\left(\frac{m}{s^2}\right)$, $V_0$ – initial velocity $\left(\frac{m}{s}\right)$.

“Starts moving” → initial velocity $= 0\frac{m}{s}$.

Then, if we put in numbers: $L = \frac{2t^2}{2} + 0t = t^2$; $t$ in seconds, $L$ in meters.

Calculate the distance covered in first 2 seconds $(L_1)$ and first 2.5 seconds $(L_2)$.

$$L_1 = 2^2 = 4 \, (m)$$ $$L_2 = 2.5^2 = 6.25 \, (m)$$

Subtract $L_1$ from $L_2$.

$$L_2 - L_1 = 6.25 - 4 = 2.25 \, (m)$$

Obtained result is nothing else, but the distance travelled in 5th half second.

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