Motion in a Plane

4.20

(a) The position is given by $\stackrel{?}{r}$ = 3. 0t ? − 2.0t 2 ? + 4.0 ?

The velocity v is given by $\stackrel{?}{v}$ = $\frac{d\stackrel{?}{r}}{dt}$ = $\frac{d}{dt}$ (3.0t ? − 2.0t 2 ? + 4.0 ?)

$\stackrel{?}{v}$ = 3.0 ? – 4.0t?

Acceleration $\stackrel{?}{a}$ = $\frac{d\stackrel{?}{v}}{dt}$ = $\frac{d}{dt}$ (3.0 ? – 4.0t?) = – 4.0?

(b) At t = 2.0s

$\stackrel{?}{v}$ = 3.0 ? – 4.0t?

The magnit

4.19

(a) The net acceleration of a particle in circular motion is always directed along the radius of the circle towards the centre (centripetal force), in the case of a circular uniform motion. Hence the statement is False.

(b) The centrifugal force direction is always along the tangent, hen

4.18 The radius of the loop, r = 1 km = 1000 m

Speed, v = 900 km/h = 250 m/s

Centripetal acceleration, AC = $\frac{v^2}{r}$ = 250 $*250$ / 1000 m/s² = 62.5 m/s² = $\frac{62.5}{9.817}$ g = 6.37g

4.17 Given, the length of the string, l = 80 cm, No. of revolution = 14, Time taken = 25 s

We know Frequency, v = $\frac{No.ofrevolution}{timetaken}$ = $\frac{14}{25}$ Hz

Angular frequency $\omega$ = 2?v = 2 $\times$ $\frac{22}{7}$ $\times \frac{14}{25}$ rad/s = 3.52 rad/s

Centripetal acceleration ac = ${\omega }^2$ r = ${3.52}^2\times \frac{80}{100}$ m/s2 = 9.91 m/s2

The direc

4.16 We know for a projectile motion, horizontal range is given by

R = $\frac{u^2}{g}sin2\theta$ , where R is the horizontal range, u is the velocity and $\theta$ is the angle of projectile

So $\frac{u^2}{g}$ = 100/sin 2 $\theta$

The cricketer will only be able to throw the ball to the maximum horizontal distance when the

4.15 The speed of the ball, u = 40 m/s

Ceiling height, h = 25m

Since the ball is thrown, it will follow the path of a projectile. For projectile motion, the maximum height for a body projected at an angle $\theta$ is given by

h= $\frac{u^2{sin}^2\theta }{2g}$ , sin2 $\theta$ = (25 $\times 2\times 9.807)/(40\times 40)$ , $\theta =33.61°$

Horizontal range is

4.14

The velocity of the boat $V_b$ = 51 km/h, the velocity of the wind $V_w$ = 72 km/h

Flag is fluttering in N-E direction, so the $V_w$ direction is N-E.

When the ship begins to sail towards North, the flag will flutter along the direction of relative velocity of the wind w.r.t. the bo

4.13 Swimming speed = 4 km/h, Width of the river = 1.0 km

Time required to cross the river = $\frac{Widthoftheriver}{Swimmingspeed}$ = $\frac{1}{4}$ h= 15 min

Speed of the river = 3 km/h

Distance covered by the river in 15 min = 3 $*\left(\frac{15}{60}\right)$ $*1000$ = 750 m

4.12 Let Vc = Velocity of the cyclist

Vr = Velocity of the rain

If V is the relative velocity, the woman must hold her umbrella in the direction of V.

V = Vr + (-Vc) = 30 + (-10) = 20 m/s

Tan $\theta$ = $\frac{Vc}{Vr}$ = $\frac{10}{30}$ , $\theta$ = 18 $°$

4.11

(a) Total distance travelled = 23 km, Total time taken = 28 min

Average speed = $\frac{Totaldistancetravelled}{Totaltimetaken}$ = $\frac{23}{28}$ km/min = 49.29 km/h

(b) Average velocity = $\frac{Totaldisplacement}{totaltimetaken}$ = $\frac{10}{28}$ km/min = 21.43 km/h