Q.A car starts from a point P at time $t = 0$ seconds and stops at point Q. The distance $x$, in metres, covered by it, in $t$ seconds is given by $x = t^2\left(2 - \dfrac{t}{3}\right)$. Find the time taken by it to reach Q and also find distance between P and Q.

Concept: Kinematics Optimization — the car stops when velocity becomes zero, so we differentiate distance to get velocity, set it to zero, and then find the corresponding distance.

Step 1: Find velocity.

$x = t^2\left(2 - \frac{t}{3}\right) = 2t^2 - \frac{t^3}{3}$

$v = \frac{dx}{dt} = 4t - t^2$

Step 2: Car stops at Q when $v = 0$.

$4t - t^2 = 0 \implies t(4 - t) = 0$

$t = 0$ (start) or $t = 4$ seconds. So time to reach Q is $4$ s.

Step 3: Distance PQ is $x$ at $t = 4$.

$x = 2(4)^2 - \frac{(4)^3}{3} = 32 - \frac{64}{3} = \frac{96 - 64}{3} = \frac{32}{3}$ m.

The problem gives distance $x$ as a function of time $t$, but the car starts from rest at P and stops at Q. That means at Q, the car’s velocity is zero again. So the key is: find when velocity is zero (other than at $t=0$), and then plug that time back into $x(t)$ to get the distance.

Let’s unpack the kinematics.

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1. Write the distance function clearly The given equation is:

Expand it:

This is a cubic in $t$, so the car speeds up, then slows down, and eventually stops.

2. Velocity is the derivative of distance

Factor it:

At $t = 0$, $v = 0$ — that’s the start at P. The car will stop again when $v = 0$ again, i.e., when $4 - t = 0$, so $t = 4$ seconds.

3. Check that $t = 4$ is indeed the stopping point

   At $t = 4$, velocity is $v = 4(4) - 4^2 = 16 - 16 = 0$. So yes, the car reaches Q at $t = 4$ seconds.

4. Find the distance PQ

   Substitute $t = 4$ into $x(t)$:

Write $32$ as $\frac{96}{3}$:

So the distance between P and Q is $\frac{32}{3}$ metres.

Method: Finding When Motion Stops (Kinematics via Derivatives)

The general approach for any "starts and stops" motion problem where position is given as a function of time.

Steps

Step 1: Recognize that "stopping" means velocity is zero

For a particle or vehicle moving along a straight path with position $x(t)$, it is momentarily at rest exactly when its velocity is zero — not when its position returns to zero. Velocity is the derivative of position:

Step 2: Differentiate the position function

Expand $x(t)$ into a polynomial in $t$ if it isn't already, then differentiate term by term to get $v(t)$.

Step 3: Solve $v(t) = 0$ and identify the relevant root

The equation $v(t)=0$ will usually include the starting instant $t=0$ as one solution (since the object starts from rest) — discard that root and take the next positive root as the stopping time.

Step 4: Substitute the stopping time back into $x(t)$

The distance travelled is simply $x(t)$ evaluated at the stopping time found in Step 3 — you are asked for a distance, not a velocity, so this final substitution is essential.

Common Mistakes

Mistake 1: Setting $x(t)=0$ instead of $v(t)=0$ to find when the car stops

Why it's wrong: $x(t)=0$ only describes when the car's position returns to its starting point, which is not what "stops" means physically. The car stops — is momentarily at rest — when its velocity is zero, a completely different condition. Correct approach: differentiate $x(t)$ to get $v(t)=\frac{dx}{dt}$ first, then solve $v(t)=0$.

Mistake 2: Reporting $t=0$ as the stopping time

Why it's wrong: solving $v(t)=0$ gives two roots here, $t=0$ and $t=4$; since the car starts from rest at $t=0$, that root describes the start, not the stop. Correct approach: discard $t=0$ (it is the given starting instant) and take the next positive root as the actual time the car reaches Q.