A body moves with a velocity of 2m/s for 5s, then its velocity uniform

1.	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.
Answer» 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

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.

Answer»

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.

A to B is uniform motion.

B to C and C to D is non-uniform motion.

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.