A body moves with a velocity of 2 ms-1 for 5 seconds, then its velocity increases uniformly to 10 ms-1 in next 5 seconds. Thereafter its
velocity begins to decrease at a uniform rate until it comes to rest after 5 seconds
1. Plot a velocity-time graph for motion of the body
2. From the graph, find the total distance covered by the body after 5 second
Answers
Answer:
Complete answer:
For the first 5s the velocity does not change with time, so the v-t graph will be parallel to the time axis and the d-t graph will be a straight line passing through the origin. Then its velocity increases uniformly for next 5s. Thus its v-t graph will be a straight line and d-t graph will be a curved line with tangent at each point giving the value of the velocity at that time. In the last 10s the velocity decreases uniformly and at last becomes zero. So the v-t graph will be a straight line with negative slope and the d-t graph will also be a curved line with negative slope
From the v-t graph we can say that motion is uniform in AB part and not uniform in BC and CD parts.
Let the initial velocity be
u.
u.
Now after
2s
2s
, as we can see from the v-t graph the distance travelled will be
S
1
=u×t=2×5=10m
S1=u×t=2×5=10m
.
Now to evaluate distance for
12s
12s
we need to find out the acceleration and retardation of the particle when the motion is not uniform.
In the point C the velocity
v=10m
s
−1
v=10ms−1
, so the acceleration in this part is given by
a=
v−u
△t
=
10 − 2
10 − 5
m
s
−2 =1.6m
s
−2
a=v−u△t=10−210−5ms−2=1.6ms−2
. Thus the distance travelled in this time interval is
S
2
=ut+
1
2
a
t
2 =2×2+(
1
2
×1.6×
5
2
)m=24
S2=ut+12at2=2×2+(12×1.6×52)m=24m
Now in the portion CD it travels with the retardation given by
r=
10−0
20−10
m
s
−2 =1m
s
−2
r=10−020−10ms−2=1ms−2
The particle travels with this retardation for last
2s
2s
of
12s
12s
. So the distance travelled in this
2s
2s
is given by
S
3
=10×2−(
1
2
×1×
2
2
)m=18m
S3=10×2−(12×1×22)m=18m
So the distance travelled in 12s is
S
1
+
S
2
+
S
3
=10+24+18=52m
S1+S2+S3=10+24+18=52m
.
In the last
10s
10s
, the particle moves with retardation. So putting the values of different quantities from the graph and from earlier calculations we get
S
4
=10×10−(
1
2
×1×
10
2
)m=50m
S4=10×10−(12×1×102)m=50m
Explanation:
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