NCERT Class 11 Physics Solutions: Chapter 4 – Motion in a Plane Part 12

Get top class preparation for CBSE right from your home: fully solved questions with step-by-step explanation- practice your way to success.

Download PDF of This Page (Size: 140K)

Question 4.31:

A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces his speed at the constant rate of 0.50 m/s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?


; 54.46° with the direction of velocity

Speed of the cyclist,

Radius of the circular turn,

Centripetal acceleration is given as:

The situation is shown in the given figure:

Figure shown the situation of example.

Figure 4.31

Suppose the cyclist begins cycling from point P and moves toward point Q. At point Q, he applies the breaks and decelerates the speed of the bicycle by. This acceleration is along the tangent at Q and opposite to the direction of motion of the cyclist.

Since the angle between is 90°, the resultant acceleration a is given by:

Where is the angle of the resultant with the direction of velocity

Question 4.32:

(a) Show that for a projectile the angle between the velocity and the x-axis as a function of time is given by

(b) Show that the projection angle for a projectile launched from the origin is given by

Where the symbols have their usual meaning.


(a) Let and respectively be the initial components of the velocity of the projectile along horizontal (x) and vertical (y) directions.

Let and respectively be the horizontal and vertical components of velocity at a point P.

Figure shown the situation of example.

Figure 4.32

Time taken by the projectile to reach point P = t

Applying the first equation of motion along the vertical and horizontal directions, we get:


(b) Maximum vertical height,

Horizontal range,

Solving equations (i) and (ii), we get: