A particle moves along a straight line so that its distance s (in metres) from a fixed point at time t (in seconds) is \(s = t^{3} - 6t^{2} + 9t\). Match List-I with List-II.
| List-I | List-II | ||
| (A) | Velocity (in m/s) at t = 2 | (I) | 9 |
| (B) | Acceleration (in m/s2) at t = 3 | (II) | −3 |
| (C) | Time (in s) at which the acceleration is zero | (III) | 6 |
| (D) | Velocity (in m/s) at t = 4 | (IV) | 2 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
The velocity is the rate of change of distance and the acceleration is the rate of change of velocity:
\( \displaystyle v = \frac{ds}{dt} = 3t^{2} - 12t + 9 \)
\( \displaystyle a = \frac{dv}{dt} = 6t - 12 \)
(A) Velocity at t = 2
This matches (II) −3.
\( v(2) = 12 - 24 + 9 = -3 \)
(B) Acceleration at t = 3
This matches (III) 6.
\( a(3) = 18 - 12 = 6 \)
(C) Time at which the acceleration is zero
This matches (IV) 2.
\( 6t - 12 = 0 \), so \( t = 2 \)
(D) Velocity at t = 4
This matches (I) 9.
\( v(4) = 48 - 48 + 9 = 9 \)