A body moves with a velocity of 2m/s for 5s, then its velocity uniformly increases to 10 m/s in the next 5s. Thereafter its velocity begins to decrease at a uniform rate until it comes to rest after 10s. Plot a v-t graph and find (a) acceleration (b) retardation and (c) total distance travelled.
Answers
Answer:
Explanation:
The body moves from A to B at velocity 2 m/s for 5 s, then it accelerates, from B to C, for 5 s and attains a velocity 10 m/s. Finally, it comes to rest in 10 s at a constant retardation, from C to D.
2.
A to B is uniform motion.
B to C and C to D is non-uniform motion.
3.
Distance travelled in 2 s
From the graph, the distance travelled in 2 s is = ut = 2 × 2 = 4 m (1)
Distance travelled in 12 s
Distance travelled in 5 s = ut = 2 × 5 = 10 m ..(2)
In the next 5 s the acceleration is, a = (v-u)/t = (10-2)/5 = 1.6 m/s 2
Distance travelled in this 5 s is, S = ut + ½ at 2
=> S = 2 × 5 + 0.5 × 1.6 × 5 2
=> S = 30 m (3)
After that the body starts slowing down. The acceleration now is = (0-10)/10 = -1 m/s 2 (negative sign indicates deceleration)
So, distance travelled in 2 s during the slowing down, S = ut + ½ at 2
=> S = 10 × 2 0.5 × 1 × 2 2
=> S = 18 m (4)
So, total distance covered in 12 s is = 10+30+18 = 58 m
Distance travelled in 20 s
Distance travelled in 20 s is equal to (2)+(3)+total distance travelled during the slowing down.
Total distance travelled during the slowing down can be found using,
v 2 = u 2 + 2aS
=> 0 = 10 2 2 × 1 × S
=> S = 50 m
So, total distance travelled in 20 s = 10+30+50 = 90 m
The body moves from A to B at velocity 2 m/s for 5 s, then it accelerates, from B to C, for 5 s and attains a velocity 10 m/s. Finally, it comes to rest in 10 s at a constant retardation, from C to D.
2.
A to B is uniform motion.
B to C and C to D is non-uniform motion.
3.
Distance travelled in 2 s
From the graph, the distance travelled in 2 s is = ut = 2 × 2 = 4 m (1)
Distance travelled in 12 s
Distance travelled in 5 s = ut = 2 × 5 = 10 m ..(2)
In the next 5 s the acceleration is, a = (v-u)/t = (10-2)/5 = 1.6 m/s 2
Distance travelled in this 5 s is, S = ut + ½ at 2
=> S = 2 × 5 + 0.5 × 1.6 × 5 2
=> S = 30 m (3)
After that the body starts slowing down. The acceleration now is = (0-10)/10 = -1 m/s 2 (negative sign indicates deceleration)
So, distance travelled in 2 s during the slowing down, S = ut + ½ at 2
=> S = 10 × 2 0.5 × 1 × 2 2
=> S = 18 m (4)
So, total distance covered in 12 s is = 10+30+18 = 58 m
Distance travelled in 20 s
Distance travelled in 20 s is equal to (2)+(3)+total distance travelled during the slowing down.
Total distance travelled during the slowing down can be found using,
v 2 = u 2 + 2aS
=> 0 = 10 2 2 × 1 × S
=> S = 50 m
So, total distance travelled in 20 s = 10+30+50 = 90 m
Answer:
Total distance travelled in 20 s = 10+30+50 = 90 m