Question

If a particle is moving in a straight line so that after \(t\) seconds its distance \(S\) (in cms) from a fixed point on the line is given by \(S = f(t) = t^3 - 5t^2 + 8t\) then the acceleration of the particle at \(t=5\) sec is (in cm/sec\(^2\))

• A. 10
• B. 30
• C. 20
• D. 40

Hint:

Distance \(\xrightarrow{d/dt}\) Velocity \(\xrightarrow{d/dt}\) Acceleration. Just differentiate twice.

Answer 1

The Correct Option is C

Step 1: Find Velocity: Velocity \(v = \frac{dS}{dt}\). \(S = t^3 - 5t^2 + 8t\) \(v = 3t^2 - 10t + 8\)
Step 2: Find Acceleration: Acceleration \(a = \frac{dv}{dt}\). \(a = \frac{d}{dt}(3t^2 - 10t + 8) = 6t - 10\)
Step 3: Calculation Process Acceleration at \(t=5\): Substitute \(t=5\) into the acceleration equation: \(a(5) = 6(5) - 10 = 30 - 10 = 20 \, \text{cm/sec}^2\).