Question

8
.
Venkat goes to the football ground to play football. The distance time graph of his
journey from his home to the ground is given as

(a) What does the graph between point B and C indicate about the motion of Venkat?
(b) Is the motion between 0 to 4 minutes uniform or non-uniform?
(c) What is the speed of Venkat between 8min-12 min?​

Answer 1

Answer:

Graph between point B and C is parallel to the x-axis as time is changing but distance remains the same means there is no motion.

Boojho is at rest position.

b) Motion between time 0 min to 4 min is nonuniform because the graph is not straight. It like two-line joining at point A.

c) At time 12-minute distance is 225 m and at time 8-minute distance is 150 m so speed is distance divided by the time taken

Speed = 225 – 150/12 – 8 = 75m./4min.

= 18.75 m/min

Step-by-step explanation:

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Answer 2

Answer:

a, Uniform Motion

b, Non Uniform

c,

Step-by-step explanation:

s=d/t

d = 75

t= 4 min

60 min = 1 h

4 min = 4/60 = 1/15

s = 75 × 15 / 1

1125 km/h

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