Motion in a Straight Line

This is a long answer type question as classified in NCERT Exemplar

let speed of two balls be V₁and V₂

Where v₁=2v and v₂=v and y₁and y₂ be the distance covered

So y₁= $\frac{v_1^2}{2g}=\frac{4v^2}{2g}$ and y₂= $\frac{v_2^2}{2g}=\frac{v^2}{2g}$

So y₁-y₂= 15

$\frac{3v^2}{2g}=15$

V²= $\sqrt[]{5*2*10}=\frac{10m}{s}$

So clearly we can say v₁=20 and v₂=10

And y₁=20m and y₂=5m

If t₂ is the time taken by ball 2 through a distance of 5m, y₂=v₂t-1/2gt²

5=10t₂-5t₂² so t₂ will be 15

Then time covered by ball 1 in 2 sec between two throws = t₁-t₂= 2-1=1s

This is a long answer type question as classified in NCERT Exemplar

(a)  for maximum velocity dv/dt=0

d/dt (6t-2t²)=0

6-4t=0 t= 6/4=1.5s

(b) v=6t-2t²

ds/dt=6t-2t²

ds=6t-2t²dt

distance in 3s, S= ${\int }_0^3\left(6t-{2t}^2\right)dt=[3t^2-\frac{2}{3}{t}^3]$ ³₀

s= 27-18=9m

average velocity = distance /time =9/3 = 3m/s

x= 6t-2t²

3=6t-2t²

After solving we get t= 9/4s approx.

(c) in periodic motion when velocity is zero

0=6t-2t²

0=t (6-2t)

So t=0, 3 sec

(d) distance covered from 0 to 3s=9m

distance covered in 3 to 6s= ${\int }_3^6\left(18-9t+t^2\right)dt$

S= (18t- $\frac{9t^2}{2}+\frac{t^3}{3}$ )⁶

S= 108-9 (18)+ $\frac{6^3}{3}-18\left(3\right)+\frac{9}{2}\left(9\right)-\frac{27}{3}$

S= -4.5m

So total distance covered = 9+ (-4.5)=4.5m

No of cycles covered in that distance =20/4.5=4.44approx

This is a long answer type question as classified in NCERT Exemplar

speed of car and truck = 72km/h = 72 (5/18) =20m/s

V= u+at

0=20+a (5) so a=-4m/s²

But retarted acceleration will be v=u+at

0=20+a (3)

So a= $\frac{20}{3}\left(t-0.5\right)$ -20/3m/s²

We also need to consider human response time = 0.5 s

V=u-at (for retarded motion)

V= 20- $\frac{20}{3}\left(t-0.5\right)$ ….1

V_t=20-4t ….2

From 1 and 2

20-=20-4t

After solving we get t= 5/4s

Distance travelled by truck in time t, S=ut+1/2at²

= 20 $*\frac{5}{4}+\frac{1}{2}\left(-4\right)*{\left(\frac{5}{4}\right)}^2=21.875m$

To avoid the bump onto the truck car must maintain distance = 23.125-21.875=1.250m

This is a long answer type question as classified in NCERT Exemplar

(a) velocity attained by a falling rain drop will be = $\sqrt[]{2gh}=\sqrt[]{2*10*1000}$

= $100\sqrt[]{2}m{s}^{-1}=\frac{510km}{h}$

(b) diameter of the rain drop = 2r=4mm

Radius = 2mm= 2 $*{10}^{-3}m$

Mass of rain drop = V $*\rho =\frac{4}{3}\pi r^3\rho =\frac{4}{3}*\frac{22}{7}*{\left(2*{10}^{-3}\right)}^3*{10}^3=3.4*{10}^{-5}kg$

Momentum of rain drop= mv= 3.4 $*{10}^5*100\sqrt[]{2}=5*{10}^{-3}kgm/s$

(c) time ,t = d/v= $\frac{4*{10}^{-3}}{100\sqrt[]{2}}=0.028*{10}^{-3}s$

(d) force exerted, F = change in momentum /time=

(e) area = $\pi R^2=\pi *{\frac{1}{2}}^2=\frac{11}{14}=0.8m^2$

number of drops striking the the umbrella with separation of 5 $*{10}^{-2}$

so net force = $\frac{0.8}{{(5*{10}^{-2})}^2}*168=53760N$

3.28 The graph has a non-uniform slope between the intervals t₁ and t₂ – the graph is not a straight line. The equations (a), (b), and (e) do not describe the motion of the particle. Only the relations (c), (d) and (f) are correct.   

3.24 The initial velocity of the ball, u = 49m/s

First Case: When the ball returns to his hands, total displacement = 0

From the relation s = ut + 0.5at², we get 0 = 49t + 0.5 (-9.81) t²

4.905 = 49t      Hence t = 10s

Second Case:

As the lift started moving up with a speed of 5 m/s, the initial velocity of the ball = 49 + 5 m/s = 54 m/s

If t' is time for the ball to return to his hand, the displacement of the ball will be = 5t'

From the relation        s = ut + 0.5 x at², we get

5t' = 54t' + 0.5 * (-9.8) t'²

49t' = 4.9 t'²

t' = 10 s