A body moves with a velocity of 2m/s for 5s then its velocity uniformely increases to 10m/s in the next 5s . Then its velocity begans to decrese at a uniform rate until it comes to rest after 10s A. Plot a velocity time graph for tje motion of tje body B. Mark the portions were motion is uniform and were motion is non uniform. C. From the graph find the total didtance moved by the body after 2s,12s and the rest 10 s

Asked by  | 27th Jul, 2013, 02:18: PM

Expert Answer:

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 (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,

= u + 2aS

=> 0 = 10 – 2 × 1 × S

=> S = 50 m

So, total distance travelled in 20 s = 10+30+50 = 90 m

Answered by  | 28th Jul, 2013, 05:53: PM

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