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The displacement and the increase in the velocity of a moving particle in the time interval of $t$ to $(t + 1)$ seconds are **125 m** and **50 m/s**, respectively. The distance travelled by the particle in $(t + 2)^{	ext{th}}$ second is ________ **m**. A). 175 B). 125 C). 100 D). 150

Edvaya · Accepted Answer

The correct answer is 175. **Solution:** Since the increase in velocity over a **1 s** interval is **50 m/s**, the constant acceleration $a =$ **50 m/s²**. Let the velocity at time $t$ be $v$. In the **1 s** interval from $t$ to $t+1$, the displacement is **125 m**. Using the equation $s = vt + \frac{1}{2}at^2$ for this 1 s interval: $125 = v(1) + \frac{1}{2}(50)(1)^2 \Rightarrow 125 = v + 25 \Rightarrow v =$ **100 m/s**. We need the distance travelled in the $(t+2)^{	ext{th}}$ second, which represents the 1 s interval spanning from $(t+1)$ to $(t+2)$. Velocity at time $(t+1)$ is $v' = v + a(1) = 100 + 50(1) =$ **150 m/s**. The displacement in this subsequent 1 s interval is: $s' = v'(1) + \frac{1}{2}a(1)^2 = 150 + \frac{1}{2}(50) = 150 + 25 =$ **175 m**.

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Physics > Mechanics > Motion In A Straight Line And Plane

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