Question

A body starting with initial velocity zero, moves in a straight line as per the law \( s = 5t^3 - 3t^2 - 5 \) (where \( s \) is distance in meters and \( t \) is time in seconds). The acceleration of the particle after 0.5 seconds will be:

• A. 1.8 m/s\(^2\)
• B. 9 m/s\(^2\)
• C. 10 m/s\(^2\)
• D. 11 m/s\(^2\)

Hint:

To find acceleration, differentiate the velocity equation with respect to time. For velocity, differentiate displacement with respect to time.

Answer 1

The Correct Option is B

Step 1: Determine velocity by differentiating the displacement equation.
The displacement equation is given as:\[s = 5t^3 - 3t^2 - 5\]Velocity \( v \) is the derivative of displacement with respect to time:\[v = \frac{ds}{dt} = \frac{d}{dt}(5t^3 - 3t^2 - 5) = 15t^2 - 6t\]Step 2: Calculate acceleration by differentiating the velocity equation.
Acceleration \( a \) is the derivative of velocity with respect to time:\[a = \frac{dv}{dt} = \frac{d}{dt}(15t^2 - 6t) = 30t - 6\]Step 3: Calculate the acceleration at \( t = 0.5 \, \text{seconds}.\)
Substitute \( t = 0.5 \) into the acceleration equation:\[a = 30(0.5) - 6 = 15 - 6 = 9 \, \text{m/s}^2\] Final Answer: \[ \boxed{9 \, \text{m/s}^2}\]