Motion in Plane

NCERT Exercise

Part 2

Question 11: A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity? Are the two equal?

Answer: Average velocity `=(10)/(28)=0.36` km per min

Average speed `=(23)/(28)=0.82` km per min

Question 12: Rain is falling vertically with a speed of 30 m s^-1. A woman rides a bicycle with a speed of 10 m s^-1 in the north to south direction. What is the direction in which she should hold her umbrella?

Answer: v_c shows the direction of woman, v_r shows the direction of rain, -v_c shows the direction of rain relative to woman’s direction. So, she must hold the umbrella tilted towards south.

Angle can be calculated as follows: tan θ `=(v_c)/(v_r)=(10)/(30)=1/3`

Or, θ = tan^-1 `1/3=18°`

Question 13: A man can swim with a speed of 4.0 km/h in still water. How long does he take to cross a river 1.0 km wide if the river flows steadily at 3.0 km/h and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?

Answer: Speed of man v_m = 4 km/h

So, time taken to cross 1 km = ¼ hr = 15 min

Distance covered with flow of river = 3 × ¼ hr = 750 m

Question 14: In a harbour, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters in the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat?

Answer: This figure shows the direction of wind v_w and direction of boat v_b. When the boat starts moving the flag will move in relative direction, i.e. v_wb

Angle between v_w and -v_b = 90° + 45°

tan β `= (51 si\n (90 + 45))/(72+51co\s(90+45))`

`=(51+co\s45)/(72-51si\n45)`

`=(51xx1/(sqrt2))/(72-51xx1/(sqrt2))`

`=(51)/(72sqrt2-51)=(51)/(50.8)=1`

So, β = 45°

So, the flag will now move towards east

Question 15: The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s^-1 can go without hitting the ceiling of the hall?

Answer: Maximum height is given by following formula:

`h=(u^2si\n^2θ)/(2g)`

Or, `25=(40^2xx\si\n^2θ)/(2xx9.8)`

Or, `si\n^2θ=(25xx19.6)/(1600)=0.30625`

Or, sin θ = 0.5534

So, θ = 33.60°

Now, horizontal range can be calculated as follows:

`R=(u^2si\n2θ)/g`

`=(40^2xxsi\n(2xx33.60°))/(9.8)`

`=(1600xx\si\n\xx67.2)/(9.8)`

`=(1600xx0.922)/(9.8)=150.53` m

Question 16: A cricketer can throw a ball to a maximum horizontal distance of 100m. How much high above the ground can the cricketer throw the same ball?

Answer: Horizontal range is given by following formula

`R=(u^2si\n2θ)/g`

We know that maximum horizontal range achieved when angle is 45°

So, `100=(u^2si\n2xx45°)/(9.8)`

Or, `980=u^2si\n90°`

Or, u² = 980

When ball is thrown vertically upward, its speed will become zero when it reaches the maximum height. Moreover, g is in the opposite direction of motion.

Using the third equation of motion

`v^2-u^2=-2gh`

Or, `-980=-2xx9.8h`

Or, `h=(980)/(2xx9.8)=50` m

Question 17: A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone?

Answer: Length of string = 80 cm = 0.8 m

Frequency ν `=(14)/(25)` Hz

Angular frequency ω = 2 π ν

`=2xx(22)/7xx(14)/(25)=(88)/(25)` rad s^-1

Centripetal acceleration `a=ω^2r`

`=((88)/(25))^2xx0.8=9.91` m s^-2

In case of circular motion, acceleration is always directed towards the centre.

Question 18: An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.

Answer: Radius of loop = 1 km = 1000 m

Speed = 900 km per hour `=900xx5/(18)=250` m/s

Acceleration `a_c=(v^2)/r`

`=(250^2)/(1000)=62.5` m s^-2

We know g = 9.8 m s^-2

Ratio `=(62.5)/(9.8)=6.38`

Or, a_c = 6.38 g

Question 19: Read each statement below carefully and state, with reasons, if it is true or false:

(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre.

Answer: False, it is true only in case of uniform circular motion.

(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point.

Answer: True, because when leaving the circular path a particle moves along a tangent to the circular path.

Answer: True, because in case of uniform circular motion, the acceleration vector keeps on changing and net resultant of all these vectors is null.

Question 20: The position of a particle is given by

r = 3.0 t i - 2.0 t² j + 4.0 k m

Where t is in seconds and coefficients have the proper units for r to be in meters.

(a) Find the v and a of the particle?

Answer: Position of the particle is given by

r = 3.0 t i - 2.0 t² j + 4.0 k m

Velocity v is given as follows:

`v=(dr)/(dt)=d/(dt)`(3.0 t i - 2.0 t² j + 4.0 k)

So, v = 3.0 t i - 4.0 t j

Acceleration a of the particle is given by

`a=(dv)/(dt)=d/(dt)(3.0i-4.0tj)`

Or, a = -4.0 j

(b) What is the magnitude and direction of velocity of the particle at t = 2.0 s?

Answer: We have velocity vector v = 3.0 i - 4.0 t j

At t = 2.0 s

v = 3.0 i - 8.0 j

The magnitude of velocity can be calculated as follows:

|v| `=sqrt(3^2+(-8)2)`

`=sqrt(73)=8.54` m/s

Direction θ = tan^-1 `(v_y)/(v_x)`

= tan^-1 `(-8)/3`

= - tan^-1 2.667 = - 69.45°

The negative sign shows that the direction of velocity is below x-axis