Motion in a Straight Line

S = Area under graph

${1 \over 2}$ $ \times $ 2$ \times $2 + 2 $ \times $ 2 + 3 $ \times $ 1 = 9 m

For both car initial speed ($\mu $) = 0

Let the acceleration of car A and car B is $a$₁ and $a$₂ respectively.

Also let the time taken to reach the finishing point for car A is t₁ and for car B is t₂.

Let at finishing point speed of car A is $v$₁ and speed of car B is $v$₂

According to the question,

t₂ $-$ t₁ = t

and   $v$₁ $-$ $v$₂ = $v$

$ \Rightarrow $  $a$₁t₁ $-$ $a$₂t₂ = $v$

$ \Rightarrow $  $a$₁t₁ $-$ $a$₂(t + t₁) = $v$ . . . . . . .(1)

As, Total distance covered by both car is equal.

So,   x_A = x_B

$ \Rightarrow $  ${1 \over 2}{a_1}t_1^2 = {1 \over 2}{a_2}t_2^2$

$ \Rightarrow $  $a$₁t$_1^2$ = $a$₂ (t + t₁)²

$ \Rightarrow $  $\sqrt {{a_1}} .{t_1}$ = $\sqrt {{a_2}} $ . (t + t₁)

$ \Rightarrow $  $\sqrt {{a_1}} .{t_1} - \sqrt {{a_2}} .{t_1} = \sqrt {{a_2}} .t$

$ \Rightarrow $  t₁ = ${{\sqrt {{a_2}} .t} \over {\sqrt {{a_1}} - \sqrt {{a_2}} }}\,\,\,\,\,\,\,.....(2)$

Now put the value of t₁ in equation (2),

($a$₁ $-$ $a$₂) t₁ $-$ $a$₂t = $v$

$ \Rightarrow $  (a₁ $-$ a₂) . ${{\sqrt {{a_2}} .t} \over {\sqrt {{a_1}} - \sqrt {{a_2}} }} - {a_2}t = v$

$ \Rightarrow $  $\left( {\sqrt {{a_1}} + \sqrt {{a_2}} } \right)\sqrt {{a_2}} .t - {a_2}t = v$

$ \Rightarrow $  $\sqrt {{a_1}{a_2}} .t + {a_2}.t - {a_2}t = v$

$ \Rightarrow $  $v$ = $\sqrt {{a_1}{a_2}} .t$

In option (A) you can see velocity versus time graph is a straight line with negative slope, so the acceleration is negative and constant. This motion is look like this.

Initially velocity is maximum and as acceleration is negative so after travelling some distance velocity will will become zero and then velocity will increase in the negative direction. At option (B) you can see same situation happens so (A) and (B) represent same situation.

From diagram, Initially at t = 0 position h = 0 and after some time h become maximum at the end h again become zero. Option (D) represent this situation.

So, A, B and D are correct.

The distance - time graph for this case will be -

But the given distance-time graph in the question does not match with this so option (C) will be wrong answer.

Till 15 sec car has accelerative motion and scooter has constant velocity in entire motion.

Total Distance travelled by the car in 15 sec, = ${1 \over 2}$ $ \times $ ${{45} \over {15}}$ $ \times $ (15)² = ${{675} \over 2}$ m

Distance travelled by scooter in 15 sec.

= V $ \times $ t = 30 $ \times $ 15 = 450 m

$\therefore\,\,\,\,$ Difference between distance travelled by the car and scooter in 15 sec,

= 450 $-$ ${{675} \over 2}$ = 112.5 m

Now, assume car catches the scooter in time t,

$\therefore\,\,\,\,$ Car travelled in t sec = scooter travelled in t sec.

$ \Rightarrow $ $\,\,\,\,$ ${{675} \over 2}$ + 45(t$-$ 15) = 30 $ \times $ t

$ \Rightarrow $ $\,\,\,\,$ 337.5 + 45t $-$ 675 = 30t

$ \Rightarrow $ $\,\,\,\,$ 15 t = 337.5

$ \Rightarrow $ $\,\,\,\,$ t = 22.5 sec.

An automobile traveling at 40 km/h can be stopped within a distance of 40 meters by using brakes. If the same automobile is traveling at 80 km/h, we need to determine the minimum stopping distance in meters, assuming no skidding occurs.

First Case

• Initial speed, $ u_1 = 40 \, \text{km/h} $


• Final speed, $ v_1 = 0 \, \text{m/s} $


• Stopping distance, $ s_1 = 40 \, \text{m} $

Using the equation:

$ v^2 - u^2 = 2as $

For the first case:

$ 0^2 - 40^2 = 2a \times 40 $

$ -1600 = 80a $

$ a = -20 \, \text{m/s}^2 $

Second Case

• Initial speed, $ u_2 = 80 \, \text{km/h} $


• Final speed, $ v_2 = 0 \, \text{m/s} $

Similarly, using the same equation:

$ 0^2 - 80^2 = 2a s_2 $

$ -6400 = 2a s_2 $

Since $ a = -20 \, \text{m/s}^2 $, divide both sides of the equation for the second case by the first case:

$ \frac{s_2}{40} = \frac{80^2}{40^2} $

$ s_2 = \frac{80 \times 80}{40} $

$ s_2 = 160 \, \text{m} $

Thus, the minimum stopping distance when the automobile is traveling at 80 km/h is 160 meters.

Acceleration of Car, a_C = 4 m/s²

Acceleration of bus, a_B = 2 m/s²

Initial distance between them, S = 200 m

Acceleration of Car with respect to bus,

a_CB = a_C $-$ a_B = 4 $-$ 2 = 2 m/s²

As initially both are at rest so, u_CB = 0

$\therefore\,\,\,$ S = U_CB $ \times $ t + ${1 \over 2}$ a_CB $ \times $ t²

$ \Rightarrow $$\,\,\,$ 200 = 0 + ${1 \over 2}$ $ \times $ 2 $ \times $ t²

$ \Rightarrow $$\,\,\,$ t² = 200

$ \Rightarrow $$\,\,\,$ t = 10$\sqrt 2 $ sec.

Given that,

acceleration (a) = $-$ C (constant)

$\therefore\,\,\,$ ${{dv} \over {dt}}$ = $-$ c

$ \Rightarrow $$\,\,\,$ ${{dv} \over {dx}}$ . ${{dx} \over {dt}}$ = $-$ c

$ \Rightarrow $$\,\,\,$ $\upsilon $ ${{dv} \over {dx}}$ = $-$ c

$ \Rightarrow $$\,\,\,$ $\int {v\,dv} $ = $-$ c$\int {dx} $

$ \Rightarrow $$\,\,\,$ ${{{v^2}} \over 2}$ = $-$ cx + k

$ \Rightarrow $$\,\,\,$ x = $-$ ${{{v^2}} \over {2c}}$ + ${k \over c}$

From this equation we can say option (c) is the correct graph.

Motion of the particle is shown below



Initially speed of the particle is maximum(v₀) and at heigth h velocity will become zero, then particle will move downward and velocity will increase gradually and become maximum when it reaches the ground. And acceleration in this entire motion will be -g, so slope of v - t will be same and negative.

So only Option (D) will be correct.

Using $h = ut + {1 \over 2}g{t^2}$

${y_1} = 10t - 5{t^2};\,\,{y_2} = 40t - 5{t^2}$

$\therefore$ $\,\,\,\,{y_2} - {y_1} = 30t\,\,\,\,$ for $\,\,\,\,t \le 8s.$

Curve will be straight line when $t \le 8s.$

when stone 1 reaches the ground then $\,\,\,{y_1} = - 240m,\,\, and\,\, t = 8s$

for $\,\,\,\,t > 8s.$

${y_2} - {y_1} = 40t + {1 \over 2}g{t^2} - 240$

So, it will be a parabolic curve till stone 2 reaches the ground. And parabola should opens upward as coefficient of t² is positive.

Time taken to reach highest point is $t = {u \over g}$

Time taken by the particle to reach the ground = $nt = {nu \over g}$

Speed on reaching ground $v = \sqrt {{u^2} + 2gH} $

Now, $v = u + at$

$ \Rightarrow \sqrt {{u^2} + 2gH} = - u + gt$

$ \Rightarrow t = {{u + \sqrt {{u^2} + 2gH} } \over g} = {{nu} \over g}$ (from question)

$ \Rightarrow 2gH = n\left( {n - 2} \right){u^2}$

Given ${{dv} \over {dt}} = - 2.5\sqrt v $

$\Rightarrow {{dv} \over {\sqrt v }} = - 2.5dt$

On integrating, $\int_{6.25}^0 {{v^{ - {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \,dv = - 2.5\int_0^t {dt} $

$ \Rightarrow \left[ {{{{v^{ + {\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}}} \over {\left( {{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}} \right)}}} \right]_{6.25}^0 = - 2.5\left[ t \right]_0^t$

$ \Rightarrow - 2{\left( {6.25} \right)^{{\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}}} = - 2.5t$

$ \Rightarrow t = 2\,sec$

For downward motion :

$v=-gt$

The velocity of the rubber ball increases in downward direction and we get a straight line between $v$ and $t$ with a negative slope.

Also applying $y - {y_0} = ut + {1 \over 2}a{t^2}$

We get $y - h = - {1 \over 2}g{t^2} \Rightarrow y = h - {1 \over 2}g{t^2}$

The graph between $y$ and $t$ is a parabola with $y=h$ at $t=0.$ As time increases $y$ decreases.

For upward motion :
The ball suffer elastic collision with the horizontal elastic plate therefore the direction of velocity is reversed and the magnitude remains the same. Here $v=u-gt$ where $u$ is the velocity just after collision. As $t$ increases, $v$ decreases. We get a straight line between $v$ and $t$ with negative slope.

Also $y = ut - {1 \over 2}g{t^2}$

All these characteristics are represented by graph $(B).$

For the body starting from rest

${x_1} = 0 + {1 \over 2}a{t^2} \Rightarrow {x_1} = {1 \over 2}a{t^2}$

For the body moving with constant speed

${x_2} = vt$

$\therefore$ ${x_1} - {x_2} = {1 \over 2}a{t^2} - vt$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow {{d\left( {{x_1} - {x_2}} \right)} \over {dt}} = at - v$

at $t=0,$ $\,\,\,\,\,\,{x_1} - {x_2} = 0$, so graph should start from origin.

For $at < v;$ the slope is negative that means ${x_1} - {x_2}$ < 0 so initially velocity of 1st body is less than second body and velocity of 1st body is increasing gradually.

For $at = v;$ the slope is zero. So ${x_1} - {x_2}$ = 0 it means here velocity of both the bodies are same.

For $at > v;$ the slope is positive. So ${x_1} - {x_2}$ > 0 it means here velocity of first body is greater than second body.

We know the relation between distance and time is.

$S = ut + {1 \over 2}a{t^2}$, which is a equation parabola. So the graph should be a parabola.

These characteristics are represented by graph $(b).$

Given that, v = v₀ + gt + ft²

We know that, $v = {{dx} \over {dt}} $

$\Rightarrow dx = v\,dt$

Integrating, $\int\limits_0^x {dx} = \int\limits_0^t {v\,dt} $

or $\,\,\,\,\,x = \int\limits_0^t {\left( {{v_0} + gt + f{t^2}} \right)} dt$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {{v_0}t + {{g{t^2}} \over 2} + {{f{t^3}} \over 3}} \right]_0^t$

or $\,\,\,\,\,x = {v_0}t + {{g{t^2}} \over 2} + {{f{t^3}} \over 3}$

At $t = 1,\,\,\,\,\,x = {v_0} + {g \over 2} + {f \over 3}.$

Given the velocity function: $v = \alpha \sqrt{x}$

We know that velocity $v$ is the rate of change of displacement with respect to time:

$\therefore \frac{dx}{dt} = \alpha \sqrt{x}$

Rearranging and separating variables, we get:

$\Rightarrow \frac{dx}{\sqrt{x}} = \alpha \, dt$

To solve this, we integrate both sides:

$\int\limits_{0}^{x} \frac{dx}{\sqrt{x}} = \alpha \int\limits_{0}^{t} dt$

This gives us:

$\Rightarrow \left[ \frac{2\sqrt{x}}{1} \right]_{0}^{x} = \alpha \left[t\right]_{0}^{t}$

Simplifying, we obtain:

$\Rightarrow 2\sqrt{x} = \alpha t $

Squaring both sides:

$\Rightarrow x = \frac{\alpha^2}{4} t^2$

Thus, the displacement $x$ varies with time $t$ as:

$x \propto t^2$

The relationship between time $ t $ and distance $ x $ is defined as $ t = ax^2 + bx $, where $ a $ and $ b $ are constants.

To find the acceleration, follow these steps:

First, differentiate the equation with respect to time $ t $:

$ \frac{d}{dt}(t) = a \frac{d}{dt}(x^2) + b \frac{dx}{dt} $

This simplifies to:

$ 1 = a \cdot 2x \frac{dx}{dt} + b \frac{dx}{dt} $

Given $ \frac{dx}{dt} = v $ (where $ v $ is the velocity), we can write:

$ 1 = 2axv + bv = v(2ax + b) $

This simplifies to:

$ 2ax + b = \frac{1}{v} $

Next, differentiate the expression $ 2ax + b = \frac{1}{v} $ again with respect to $ t $:

$ 2a \frac{dx}{dt} + 0 = -\frac{1}{v^2} \frac{dv}{dt} $

Using $ \frac{dx}{dt} = v $, we get:

$ 2av = -\frac{1}{v^2} \frac{dv}{dt} $

Solving for $ \frac{dv}{dt} $:

$ \frac{dv}{dt} = -2a v^3 $

Since acceleration $ f = \frac{dv}{dt} $, we have:

$ f = -2av^3 $

The velocity of parachutist when parachute opens at 50 m is

$u = \sqrt {2gh} = \sqrt {2 \times 9.8 \times 50} = \sqrt {980} $

The velocity at ground, $v=3m/s$

$\therefore$ $S = {{{v^2} - {u^2}} \over {2 \times \left( { - 2} \right)}} = {{{3^2} - 980} \over { - 4}} \approx 243\,m$

Initially he has fallen $50$ $m.$

$\therefore$ Total height from where

He bailed out $=243+50=293m$

Initially car starts from rest so u = 0.

Now distance from $A$ to $B$,

$\,\,\,\,\,\,\,\,\,\,$ $ S = {1 \over 2}ft_1^2 $

$\Rightarrow ft_1^2 = 2S$

Distance from $B$ to $C$ $ = \left( {f{t_1}} \right)t$

In B to C velocity is constant and v = ${f{t_1}}$

Distance from $C$ to $D$
$\,\,\,\,\,\,\,\,\,\,$ $ = {{{u^2}} \over {2a}} = {{{{\left( {f{t_1}} \right)}^2}} \over {2\left( {f/2} \right)}} = ft_1^2 = 2S$

$ \Rightarrow S + f\,{t_1}t + 2S = 15S $

$\Rightarrow f\,{t_1}t = 12S$ ........(1)

But$\,\,\,\,\,\,\,\,\,$ ${1 \over 2}f\,t_1^2 = S$ .........(2)

On dividing the above two equations, we get ${t_1} = {t \over 6}$

$ \Rightarrow S = {1 \over 2}f{\left( {{t \over 6}} \right)^2} = {{f\,{t^2}} \over {72}}$

We know that equation of motion, $s = ut + {1 \over 2}g{t^2},\,\,$

Initial speed of ball is zero and it take T second to reach the ground.

$\therefore$ $h = {1 \over 2}g{T^2}$

After $T/3$ second, vertical distance moved by the ball

$h' = {1 \over 2}g{\left( {{T \over 3}} \right)^2} $

$\Rightarrow h' = {1 \over 2} \times {{8{T^2}} \over 9}$

$ = {h \over 9}$

$\therefore$ Height from ground

$ = h - {h \over 9} = {{8h} \over 9}$

Assume $a$ be the retardation for both the vehicle then

In case of automobile,

$u_1^2 - 2a{s_1} = 0$

$ \Rightarrow u_1^2 = 2a{s_1}$

And in case for car,

$u_2^2 = 2a{s_2}$

$\therefore$ ${\left( {{{{u_2}} \over {{u_1}}}} \right)^2} = {{{s_2}} \over {{s_1}}}$

$ \Rightarrow {\left( {{{120} \over {60}}} \right)^2} = {{{s_2}} \over {20}}$

$ \Rightarrow$ s₂ = 80 m

Given that $x = \alpha {t^3}\,\,\,\,$ and $\,\,\,\,y = \beta {t^3}$

$\therefore$ ${v_x} = {{dx} \over {dt}} = 3\alpha {t^2}\,\,\,\,$

and$\,\,\,\,\,{v_y} = {{dy} \over {dt}} = 3\beta {t^2}$

$\therefore$ $v = \sqrt {v_x^2 + v_y^2} $

$ = \sqrt {9{\alpha ^2}{t^4} + 9{\beta ^2}{t^4}} $

$ = 3{t^2}\sqrt {{\alpha ^2} + {\beta ^2}} $

For case 1 :
u = 50 km/hr = ${{50 \times 1000} \over {3600}}$ m/s = ${{125} \over 9}$ m/s, v = 0, s = 6 m, $a$ = ?

$\therefore$ 0² = u² + 2$a$s

$ \Rightarrow $ $a = - {{{u^2}} \over {2s}}$

$ \Rightarrow $ $a = - {{{{\left( {{{125} \over 9}} \right)}^2}} \over {2 \times 6}}$ = $-$16 m/s²

For case 2 :
u = 100 km/hr = ${{100 \times 1000} \over {3600}}$ m/s = ${{250} \over 9}$ m/s, v = 0, $a$ = $-$16, s = ?

$\therefore$ 0² = u² + 2$a$s

$ \Rightarrow $ $s = - {{{u^2}} \over {2a}}$

$ \Rightarrow $ $s = - {{{{\left( {{{250} \over 9}} \right)}^2}} \over {2 \times -16}}$ = 24 m

Assume the initial velocity of each particle is = u

And height of building = h

If final velocity of A is v_A then v_A² = u² + 2(-g)(-h) = u² + 2gh

If final velocity of B is v_B then v_B² = u² + 2gh

$\therefore$ v_A = v_B

Sign Rule : Take the direction of initial velocity positive opposite direction as negative.

Here for ball A initial velocity u is upward so upward is positive and downward is negative. That is why gravity is = - g and height = - h

And for ball B initial velocity u is downward so downward is positive and upward is negative. That is why gravity is = + g and height = + h

Given the initial speeds of two identical cars as $u$ and $4u$, and considering that both cars eventually stop (final speed $v = 0$), we note that both cars decelerate with the same acceleration $-a$.

Using the kinematic equation:

$ v^2 = u^2 - 2as $

Since $v = 0$,

$ 0 = u^2 - 2as $

Hence,

$ u^2 = 2as $

For the first car with speed $u$,

$ u^2 = 2a s_1 \quad \text{...(i)} $

For the second car with speed $4u$,

$ (4u)^2 = 2a s_2 \quad \text{...(ii)} $

Dividing equation (i) by equation (ii),

$ \frac{u^2}{(4u)^2} = \frac{2a s_1}{2a s_2} $

$ \frac{u^2}{16u^2} = \frac{s_1}{s_2} $

$ \frac{1}{16} = \frac{s_1}{s_2} $

Thus, the ratio of the stopping distances of the two cars is $\frac{1}{16}$.