(a) A particle moves along a line according to the law s(t)=2t³-9t²+12t-4, where t≥0. (i) At what times the pa

Question 148 of 148

s(t)=2t^3-9t^2+12t-4

(i) At what times the particle changes direction ?

(ii) Find the total distance travelled by the particle in the first 4 seconds.

(iii) Find the particle's acceleration each time the velocity is zero. OR

(b) Find the non-parametric form of Vector equation and Cartesian equation of the plane passing through the point

(1, -2, 4)

and perpendicular to the plane

x+2y-3z=11

and parallel to the line

\dfrac{x+7}{3}=\dfrac{y+3}{-1}=\dfrac{z}{1}

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2026Subjective· 5mImportance★★★★★

100% · 148/148 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Meaning of Derivative and Rate of Change

The derivative $f'(x)$ of a function carries two equivalent readings, and this chapter leans on both throughout.

As a slope. For the curve $y=f(x)$ , the slope of the chord joining $(x,f(x))$ and $(x+h,f(x+h))$ is the Newton quotient

\frac{f(x+h)-f(x)}{h}.

Taking $h\to0$ gives the slope of the curve at $(x,f(x))$ itself:

f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}.

If $\theta$ is the angle the tangent makes with the positive $x$ -axis (measured anticlockwise), then $f'(x)=\tan\theta$ .

As a rate of change. $f'(x)=\dfrac{dy}{dx}$ is also the instantaneous rate of change of $y$ with respect to $x$ ; over an interval $[a,b]$ the average rate of change is the ordinary difference quotient $\dfrac{f(b)-f(a)}{b-a}$ (a chord slope), while the derivative at a single point is the instantaneous rate.

Motion along a line. If $s=f(t)$ is the position of an object at time $t$ (measured from a fixed origin, positive direction = forward):

v(t)=\frac{ds}{dt},\qquad a(t)=\frac{dv}{dt}=\frac{d^2s}{dt^2}.

Speed $=|v(t)|=\left|\dfrac{ds}{dt}\right|$ — always non-negative, regardless of direction.
$v(t)=0$ : the particle is momentarily at rest.
$v(t)>0$ : moving forward; $v(t)<0$ : moving backward.
The particle changes direction exactly where $v(t)$ changes sign (not merely where it is zero — the sign must flip on either side).
If the particle reverses direction at time $t_c\in(t_1,t_2)$ , the total distance travelled from $t_1$ to $t_2$ is $|s(t_c)-s(t_1)|+|s(t_2)-s(t_c)|$ — NOT simply $|s(t_2)-s(t_1)|$ , since backtracking would otherwise cancel out.
Near Earth's surface a freely falling body has constant acceleration $g\approx9.8\ \text{m/s}^2$ ( $32\ \text{ft/s}^2$ ), giving $a=-g,\ v=-gt+v_0,\ s=-\tfrac12gt^2+v_0t+s_0$ . …

Start your 14-day free trial to unlock the full solution →

(a) Uses the velocity $v(t)=s'(t)$ to locate direction reversals, then breaks the total distance into monotone pieces and evaluates acceleration at the zero-velocity instants; (b) builds the required plane's normal as the cross product of the given plane's normal and the line's direction, since both must lie in it. Both alternatives answered below.

(a) Motion of the particle $s(t)=2t^3-9t^2+12t-4$

1. Velocity.

$v(t)=s'(t)=6t^2-18t+12=6(t^2-3t+2)=6(t-1)(t-2)$

2. (i) When the particle changes direction. $v(t)=0$ at $t=1$ and $t=2$ . Check the sign of $v(t)=6(t-1)(t-2)$ on each side:

$0\le t<1$ : both factors negative $\Rightarrow v>0$ (moving forward).
$1<t<2$ : $(t-1)>0,\ (t-2)<0\Rightarrow v<0$ (moving backward).
$t>2$ : both factors positive $\Rightarrow v>0$ (moving forward again).

Since $v$ genuinely changes sign at both points, the particle reverses direction at $t=1$ s and $t=2$ s.

3. (ii) Total distance in $[0,4]$ . Evaluate the position at each turning instant and the endpoints:

$s(0)=-4,\quad s(1)=2-9+12-4=1,\quad s(2)=16-36+24-4=0,\quad s(4)=128-144+48-4=28$

Distance travelled on each monotone piece:

$|s(1)-s(0)|=|1-(-4)|=5,\qquad|s(2)-s(1)|=|0-1|=1,\qquad|s(4)-s(2)|=|28-0|=28$

Total distance $=5+1+28=\mathbf{34}$ units (note this is more than $|s(4)-s(0)|=32$ , since the particle backtracks).

4. (iii) Acceleration when velocity is zero.

$a(t)=v'(t)=12t-18$

At $t=1$ : $a(1)=12-18=-6$ . At $t=2$ : $a(2)=24-18=6$ .

(b) Plane through $(1,-2,4)$ , perpendicular to $x+2y-3z=11$ , parallel to $\dfrac{x+7}3=\dfrac{y+3}{-1}=\dfrac z1$

1. Data. Point $A=(1,-2,4)$ . Normal of the given plane: $\vec n_1=(1,2,-3)$ — since our plane is perpendicular to this plane, $\vec n_1$ is perpendicular to our plane's normal, i.e. $\vec n_1$ lies in our plane. Direction of the given line: $\vec d=(3,-1,1)$ — since our plane is parallel to this line, $\vec d$ also lies in our plane.

2. Normal of the required plane. Since $\vec n_1$ and $\vec d$ both lie in the plane, its normal is …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2020Set ANNUAL1 markMCQ
$s(t) = 3t^2 - 2t - 8$
(a) $t=3$
(b) $t=0$
(c) $t=\dfrac{1}{3}$
(d) $t=1$
›Reveal solutionSolution
Setting the velocity $v(t)=\dfrac{ds}{dt}$ to zero gives the time at which the particle is at rest, $t=\dfrac13$ .

The position function is $s(t)=3t^2-2t-8$ .

The velocity is the first derivative of position with respect to time: $v(t)=\dfrac{ds}{dt}=6t-2$ . …
Start your 14-day free trial to see the full solution →
CBSE 2019Set ANNUAL1 markMCQ
$Q. The surface area of a sphere when the volume is increasing at the same rate as its radius, is :$
(a) $4\pi$
(b) $\dfrac{4\pi}{3}$
(c) $1$
(d) $\dfrac{1}{2\pi}$
›Reveal solutionSolution
When the rate of change of volume equals the rate of change of radius, the sphere's surface area equals $1$ .

Volume of a sphere: $V=\dfrac{4}{3}\pi r^3$ .

Differentiate with respect to time $t$ : $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$ .

We are given $\dfrac{dV}{dt}=\dfrac{dr}{dt}$ (the volume increases at the same rate as the radius).

Substitute: $\dfrac{dr}{dt}=4\pi r^2\dfrac{dr}{dt}$ . …
Start your 14-day free trial to see the full solution →
CBSE 2017Set ANNUAL1 markMCQ
$y = F(t)$
(a) Gradient of the velocity/time graph
(b) Gradient of the distance/time graph
(c) Gradient of the acceleration/time graph
(d) Gradient of the velocity/distance graph
›Reveal solutionSolution
Since velocity is $dy/dt$ and acceleration is $dv/dt$ , acceleration is the gradient of the velocity–time graph.

Given $y=F(t)$ is displacement (distance) as a function of time $t$ .

Velocity $v=\dfrac{dy}{dt}$ is, by definition, the gradient (slope) of the distance–time graph.

Acceleration $a=\dfrac{dv}{dt}$ is the rate of change of velocity with respect to time — i.e., the gradient of the velocity–time graph. …
Start your 14-day free trial to see the full solution →

←Previous questionQ62 — If limθ→0((1-cos mθ)/(1-cos nθ))=1, then prove that m=± n →Next chapterCh 8Differentials and Partial Derivatives

Q.(a) A particle moves along a line according to the law s(t)=2t3−9t2+12t−4s(t)=2t^3-9t^2+12t-4s(t)=2t3−9t2+12t−4, where t≥0t\ge0t≥0.

(a) Motion of the particle s(t)=2t3−9t2+12t−4s(t)=2t^3-9t^2+12t-4s(t)=2t3−9t2+12t−4

(b) Plane through (1,−2,4)(1,-2,4)(1,−2,4), perpendicular to x+2y−3z=11x+2y-3z=11x+2y−3z=11, parallel to x+73=y+3−1=z1\dfrac{x+7}3=\dfrac{y+3}{-1}=\dfrac z13x+7​=−1y+3​=1z​

🎓Unlock everything free for 14 days

Q.(a) A particle moves along a line according to the law $s(t)=2t^3-9t^2+12t-4$ , where $t\ge0$ .

(a) Motion of the particle $s(t)=2t^3-9t^2+12t-4$

(b) Plane through $(1,-2,4)$ , perpendicular to $x+2y-3z=11$ , parallel to $\dfrac{x+7}3=\dfrac{y+3}{-1}=\dfrac z1$

Unlock everything free for 14 days