A body moving along x-axis such that x(t)=(t-2)square . Calculate distance covered by the body in 4 seconds .
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Answered by
2
Answer:
solved
Explanation:
X (t) = (T - 2)^2
X (t) = (t - 2)²
X (4) = (4 - 2)² = 2² = 4
v = 2(t - 2)
Answered by
7
Given distance-time relation
x=(t2−4t+6)
Velocity
v=dxdt=2t−4
Acceleration
a=dvdt=2ms−2
Velocity at
t
=
0
v
0
=
−
4
m
s
−
1
It can be seen that the body was having its velocity in
−
v
e
direction from initial location of
6
m
. Due to positive acceleration the velocity increased finally to
2
m
s
−
1
through
0
m
s
−
1
Time when velocity
=
0
Using the kinematic expression
v=u+at
0=−4+2t
⇒
t=2s
Distance moved in these
2s
v2−u2=2as(0→2) (−4)2−02=2×2
s(0→2)
⇒s(0→2)=164=4m
Velocity at
t=3s
v3=2×3−4
v3=2ms−1
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