A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 4.21. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist ?
A cyclist starts from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in Fig. 4.21. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist ?
Given: The radius of the circular park is 1 km, time taken to complete the round trip is 10 min.
The circular path and the path of the cyclist is shown below:
(a)
Asthe initial and the final pointscoincide at the centre of the circle therefore, the net displacement of the cyclist is zero.
(b)
The average velocity is the ratio of net displacement to the total time taken during the motion.
Since,the net displacement of the cyclist is zero therefore, the average velocity is also zero.
(c)
The length of the arc PQ is given as,
PQ=θ×R
where, θ is the angular displacement and R is the radius of the park.
By substituting the given values in the above equation, we get
PQ= 1 /4 ×2π×1 km
The average speed is given as,
v= Total distance travelled /Total time taken = OP+ PQ+QO/ t ………….(1)
time taken, t = 10 min = 10/60 hr = 1/6 hr
By substituting the given values in equation (1), we get
v= (1 km+( 1/4 ×2π×1 km )+1 km)/ (1/ 6 h)  =6( 2+ 1.57) km h -1 ≈21.4 km h-1
Thus, the average speed of the cyclist is 21.4 km h-1.
Given: The radius of the circular park is 1 km, time taken to complete the round trip is 10 min.
The circular path and the path of the cyclist is shown below:
(a)
Asthe initial and the final pointscoincide at the centre of the circle therefore, the net displacement of the cyclist is zero.
(b)
The average velocity is the ratio of net displacement to the total time taken during the motion.
Since,the net displacement of the cyclist is zero therefore, the average velocity is also zero.
(c)
The length of the arc PQ is given as,
PQ=θ×R
where, θ is the angular displacement and R is the radius of the park.
By substituting the given values in the above equation, we get
PQ= 1 /4 ×2π×1 km
The average speed is given as,
v= Total distance travelled /Total time taken = OP+ PQ+QO/ t ………….(1)
time taken, t = 10 min = 10/60 hr = 1/6 hr
By substituting the given values in equation (1), we get
v= (1 km+( 1/4 ×2π×1 km )+1 km)/ (1/ 6 h)  =6( 2+ 1.57) km h -1 ≈21.4 km h-1
Thus, the average speed of the cyclist is 21.4 km h-1.
