Question

For a particle moving along a straight line path, the displacements in third and fifth seconds of its motion are 10 m and 18 m respectively. The speed of the particle at time t=4s is

• A. 32 ms$^{-1}$
• B. 8 ms$^{-1}$
• C. 12 ms$^{-1}$
• D. 16 ms$^{-1}$

Hint:

The formula for displacement in the $n^{th}$ second, $s_n = u + a(n-1/2)$, is very useful for problems involving specific time intervals. It is derived from the difference in positions at time $n$ and $n-1$: $s(n) - s(n-1)$.

Answer 1

The Correct Option is D

Step 1: Use the formula for displacement in the $n^{th$ second.}
The displacement, $s_n$, of a particle undergoing uniformly accelerated motion during the $n^{th}$ second is given by: \[ s_n = u + \frac{a}{2}(2n-1). \] where $u$ is the initial velocity and $a$ is the constant acceleration.

Step 2: Set up equations using the given data.
For the third second ($n=3$), the displacement is 10 m. \[ s_3 = u + \frac{a}{2}(2(3)-1) = u + \frac{5a}{2} = 10. (1) \] For the fifth second ($n=5$), the displacement is 18 m. \[ s_5 = u + \frac{a}{2}(2(5)-1) = u + \frac{9a}{2} = 18. (2) \]

Step 3: Solve the system of equations for $u$ and $a$.
Subtract Equation (1) from Equation (2): \[ \left(u + \frac{9a}{2}\right) - \left(u + \frac{5a}{2}\right) = 18 - 10. \] \[ \frac{4a}{2} = 8 \implies 2a = 8 \implies a = 4 \text{ m/s}^2. \] Substitute the value of $a$ back into Equation (1): \[ u + \frac{5(4)}{2} = 10 \implies u + 10 = 10 \implies u = 0 \text{ m/s}. \]

Step 4: Calculate the speed at t=4s.
The speed (velocity) $v$ at time $t$ is given by the kinematic equation $v = u + at$. We need to find the speed at $t=4$s. \[ v(4) = 0 + (4)(4) = 16 \text{ m/s}. \] \[ \boxed{v(4) = 16 \text{ m/s}}. \]