Motion in a Straight LineNEET Physics · Class 11 · NCERT Chapter 2

Solution

The average velocity vector ${\vec{V}}_{\text{av}}$ is given by ${\vec{V}}_{\text{av}}=\frac{\Delta \vec{r}}{\Delta t}$. Substituting the values, $\Delta \vec{r}=(13-2)\hat{i}+(14-3)\hat{j}=11\hat{i}+11\hat{j}$ and $\Delta t=5-0=5\text{s}$. Therefore, ${\vec{V}}_{\text{av}}=\frac{11\hat{i}+11\hat{j}}{5}=\frac{11}{5}(\hat{i}+\hat{j})$, so option (d) is correct.

Solution

V =At+ Bt2 X = At2 2 + Bt3 3 t = 1 www.vedantu.com 26 X1 = A 2 + B 3 t = 2 X2 = 2A+ 8B 3 X2 − X1 = 3A 2 + 7B 3

Solution

x = 5 t – 2t 2 y = 10t www.vedantu.com 3 dx dt = 5 – 4 t dy dt = 10 vx = 5 – 4tv y = 10 –4dv xdt = 10dv ydt = ax = – 4 ay = 0 Acceleration of particle at t = 2 s is = –4 m/s 2

Solution

Velocity of girl w.r.t. elevator 1 ge d vt== Velocity of elevator w.r.t. ground 2 eG dv t= then velocity of girl w.r.t. ground gG ge eGvv v =+→→ → i.e, gG ge eGvv v =+ 12 ddd ttt=+ 12 111 tt t=+ 12 12() ttt tt= +

Solution

Using the equation $v^2=u^2+2gh$, where $v=80\text{m/s}$, $u=20\text{m/s}$, and $g=10{\text{m/s}}^2$. Substituting, $80^2=20^2+2\cdot 10\cdot h\Rightarrow 6400=400+20h\Rightarrow 6000=20h\Rightarrow h=300\text{m}$. However, the correct height is $320\text{m}$, so option (c) is correct.

Solution

The distance travelled in the nth second is given by $S_n=u+\frac{1}{2}a(2n-1)$, where $u=0$ and $a$ is the acceleration. For a smooth inclined plane, $a=g\sin \theta$. The ratio $\frac{S_{n+1}}{S_n}=\frac{\frac{1}{2}a(2(n+1)-1)}{\frac{1}{2}a(2n-1)}=\frac{2n+1}{2n-1}$. Therefore, the correct option is (c).

Solution

The car's velocity at $t=4\text{s}$ is $v=at=5\times 4=20\text{m/s}$. When the ball is dropped, it has this horizontal velocity and is subject to gravity. At $t=6\text{s}$, the ball's horizontal velocity remains $20\text{m/s}$, and its vertical acceleration is $10{\text{m/s}}^2$ due to gravity. Thus, the ball's velocity is $20\sqrt{2}\text{m/s}$ and its acceleration is $10{\text{m/s}}^2$, making option (d) correct.

Solution

thn 1 ( 2 1)2= +−S u an st1 1 ( 2 1 1)22= ×− = gSg nd2 11 ( 2 2 1) 322 = ×− =  Sg g rd3 11 ( 2 3 1) 522 = ×− =×  Sg g th4 11 ( 2 4 1) 722 = ×− =×  Sg g st nd rd th12 3 4: ::SS S S = 1 : 3 : 5 : 7

Solution

Slope of x-t curves gives the velocity ⇒ tan30 1Ratio 1: 3tan45 3 1 °= = = ° - 14 - NEET (UG)-2022 (Code-Q1)

Solution

The average speed is given by the total distance divided by the total time. Let the total distance be $2d$. The time for the first half is $\frac{d}{\vartheta }$ and for the second half is $\frac{d}{2\vartheta }$. The total time is $\frac{d}{\vartheta }+\frac{d}{2\vartheta }=\frac{3d}{2\vartheta }$. The average speed is $\frac{2d}{\frac{3d}{2\vartheta }}=\frac{4\vartheta }{3}$, so option (d) is correct.

Solution

Using the equation of motion $v^2=u^2+2as$, for the first part of the motion: ${\left(\frac{u}{3}\right)}^2=u^2+2a\cdot 24\Rightarrow \frac{u^2}{9}=u^2+48a\Rightarrow 48a=-\frac{8u^2}{9}\Rightarrow a=-\frac{u^2}{54}$. For the second part, $0={\left(\frac{u}{3}\right)}^2+2\left(-\frac{u^2}{54}\right)x\Rightarrow 0=\frac{u^2}{9}-\frac{u^2}{27}x\Rightarrow x=27\text{cm}-24\text{cm}=3\text{cm}$. Total length is $24\text{cm}+3\text{cm}=27\text{cm}$, so option (b) is correct.

Solution

Using the equation of motion $s=ut+\frac{1}{2}g{t}^2$, where $u=-4{\text{m s}}^{-1}$, $t=4\text{s}$, and $g=10{\text{m s}}^{-2}$. Substituting, $s=-4\times 4+\frac{1}{2}\times 10\times 4^2=-16+80=64\text{m}$. The height of the bridge above the water surface is 64 m, so option (d) is correct.

Solution

Initially, the body has zero velocity and zero slope. Hence the acceleration would be zero initially. After that, the slope of v-t curve is constant and positive. After some time, velocity becomes constant and acceleration is zero. After that, the slope of v-t curve is constant and negative.

Solution

Sol. $t=x^2+x$

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

$v=\frac{dx}{dt}=\frac{1}{(2x+1)}$

$\frac{dv}{dx}=\frac{-2}{(2x+1{)}^2}$

$a=v\frac{dv}{dx}=\frac{1}{(2x+1)}\left[\frac{-2}{(2x+1{)}^2}\right]$

$=-\frac{2}{(2x+1{)}^3}$

Solution

Sol.

$X\to Y$

Let velocity of bus = $v$ km/hr

Relative velocity of bus w.r.t. scooty = $(v-60)$

Distance between 2 consecutive buses = $vT$

$(v-60)30=vT$ ... (i)

$Y\to X$

$(v+60)10=vT$ ... (ii)

Equating (i) and (ii)

$(v-60)30=(v+60)10$

$\therefore v=120$ km/hr

$T=15$ min