Question

A car is started 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 \( x = t^2(2 - \frac{t}{3}) \). Find the time required by the car to reach at point Q and also find the distance between P and Q.

Hint:

Remember the relationship between displacement (x), velocity (v), and acceleration (a): \( v = \frac{dx}{dt} \) and \( a = \frac{dv}{dt} = \frac{d^2x}{dt^2} \). The word "stopped" or "at rest" always implies that the velocity is zero.

Answer 1

Step 1: Understanding the Concept:
The problem describes the motion of a car with its distance from the start point given as a function of time. The car stopping at point Q means its velocity becomes zero at that point in time. We need to use calculus to find the velocity from the distance function.
Step 2: Key Formula or Approach:
1. Express the distance \(x\) as a polynomial of \(t\).
2. Velocity \(v\) is the first derivative of distance with respect to time: \( v = \frac{dx}{dt} \).
3. Set the velocity \(v = 0\) to find the time \(t\) when the car stops.
4. Substitute this value of \(t\) back into the distance equation to find the total distance covered.
Step 3: Detailed Explanation:
The distance function is given by: \[ x(t) = t^2\left(2 - \frac{t}{3}\right) = 2t^2 - \frac{t^3}{3} \] To find the velocity, we differentiate \(x(t)\) with respect to \(t\): \[ v(t) = \frac{dx}{dt} = \frac{d}{dt}\left(2t^2 - \frac{t^3}{3}\right) \] \[ v(t) = 4t - \frac{3t^2}{3} = 4t - t^2 \] The car stops at point Q, so its velocity is zero. We set \(v(t) = 0\): \[ 4t - t^2 = 0 \] \[ t(4 - t) = 0 \] This gives two possible times: \( t = 0 \) and \( t = 4 \).
\( t = 0 \) corresponds to the starting point P. Therefore, the time required for the car to reach point Q is \( t = 4 \) seconds.
Now, we find the distance between P and Q by substituting \( t = 4 \) into the distance equation: \[ x(4) = 2(4)^2 - \frac{(4)^3}{3} \] \[ x(4) = 2(16) - \frac{64}{3} \] \[ x(4) = 32 - \frac{64}{3} = \frac{96 - 64}{3} = \frac{32}{3} \] The distance between P and Q is \( \frac{32}{3} \) metres.
Step 4: Final Answer:
The time required to reach point Q is 4 seconds, and the distance between P and Q is \( \frac{32}{3} \) metres.