A body moves with a velocity of 2m/s for 5 seconds . It’s
velocity then increases uniformly for 5 seconds, to
10m/s. Thereafter, it comes to rest in 10 seconds with
uniform decrease in velocity.
1.Plot a velocity time-graph for the motion of the body.
[5marks]
2.On the graph mark the portions when the body is
travelling with uniform motion. [1mark]
3.Which parts of graph represents non-uniform
motion? [1mark]
4.Calculate the acceleration of the body. [1mark]
5.From the graph find the total distance moved by the
body after 10 seconds.[2marks]
Answers
Answer:
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vanshika9270
19.06.2018
Science
Secondary School
+13 pts
Answered
A body moves with a velocity of 2m/s for 5s, then its velocity uniformly increases to 10m/s in next 5s. there after its velocity begins to decrease at a uniform rate until it comes to rest after 10s.
plot a velocity-time graph for the motion of the body.
mark the position of the graph to sjow when the motion of the body is uniform and when it is non-uniform.
from the graph find the total distance moved, by the body in 2s, 12s and in the lat 20s.
2
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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/s2
Distance travelled in this 5 s is, S = ut + ½ at2
=> S = 2 × 5 + 0.5 × 1.6 × 52
=> S = 30 m …………………(3)
After that the body starts slowing down. The acceleration now is = (0-10)/10 = -1 m/s2 (negative sign indicates deceleration)
So, distance travelled in 2 s during the slowing down, S = ut + ½ at2
=> S = 10 × 2 – 0.5 × 1 × 22
=> 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,
v2 = u2 + 2aS
=> 0 = 102 – 2 × 1 × S
=> S = 50 m
So, total distance travelled in 20 s = 10+30+50 = 90 m
Explanation:
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