Question

A car is started to move from a point P at time t = 0 and is stopped at the point Q. The distance x metre covered by the car in t second is given by \(x = t^2\left(3-\frac{t}{2}\right)\). Find the time required by the car to reach at the point Q and also find the distance between P and Q.

Hint:

In kinematics problems, "starts from rest" and "stops" both imply that the velocity is zero at that specific time. "Starts at the origin" means the initial position is zero. Carefully translate these phrases into mathematical conditions.

Answer 1

Step 1: Understanding the Concept:
The problem describes the motion of a car. The key information is that the car "is stopped" at point Q. In physics and calculus, this means that the car's instantaneous velocity is zero at the time it reaches Q. Velocity is the first derivative of the distance function with respect to time.
Step 2: Key Formula or Approach:
1. The distance function is given: \(x(t) = t^2(3 - \frac{t}{2})\). 2. The velocity function is the derivative of the distance function: \(v(t) = \frac{dx}{dt}\). 3. Set \(v(t) = 0\) to find the time 't' when the car stops. 4. Substitute this value of 't' back into the distance function \(x(t)\) to find the total distance covered.
Step 3: Detailed Explanation:
First, expand the distance function for easier differentiation: \[ x(t) = 3t^2 - \frac{t^3}{2} \] Now, find the velocity function \(v(t)\) by differentiating \(x(t)\) with respect to t: \[ v(t) = \frac{dx}{dt} = \frac{d}{dt}\left(3t^2 - \frac{t^3}{2}\right) \] \[ v(t) = 6t - \frac{3t^2}{2} \] The car stops at point Q, so its velocity is 0 at that time. We set \(v(t) = 0\): \[ 6t - \frac{3t^2}{2} = 0 \] Factor out the common term \(t\): \[ t\left(6 - \frac{3t}{2}\right) = 0 \] This gives two possible solutions for t: \[ t = 0 \quad \text{or} \quad 6 - \frac{3t}{2} = 0 \] \(t=0\) corresponds to the starting point P, where the car was at rest. The other solution gives the time to reach point Q. \[ 6 = \frac{3t}{2} \] \[ 12 = 3t \] \[ t = 4 \] So, the time required for the car to reach point Q is 4 seconds.
Now, to find the distance between P and Q, we substitute \(t=4\) into the distance function \(x(t)\): \[ x(4) = 3(4)^2 - \frac{(4)^3}{2} \] \[ x(4) = 3(16) - \frac{64}{2} \] \[ x(4) = 48 - 32 = 16 \] The distance between P and Q is 16 metres.
Step 4: Final Answer:
The time required to reach point Q is 4 seconds, and the distance between P and Q is 16 metres.