(ii)A body moving along X- axis such that x(t) = (t3-4t+6)m., calculate distance covered by the body in 4
seconds. Instantaneous velocity at t= 3s (5)
Answers
Answer:
x
=
(
t
2
−
4
t
+
6
)
Velocity
v
=
d
x
d
t
=
2
t
−
4
Acceleration
a
=
d
v
d
t
=
2
m
s
−
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
+
a
t
0
=
−
4
+
2
t
⇒
t
=
2
s
Distance moved in these
2
s
v
2
−
u
2
=
2
a
s
(
0
→
2
)
(
−
4
)
2
−
0
2
=
2
×
2
s
(
0
→
2
)
⇒
s
(
0
→
2
)
=
16
4
=
4
m
Velocity at
t
=
3
s
v
3
=
2
×
3
−
4
v
3
=
2
m
s
−
1
Distance moved from
t
=
2
to
t
=
3
when velocity changed from
0
m
s
−
1
to
2
m
s
−
1
, movement was also became in the positive direction.
2
2
−
0
2
=
2
×
2
×
s
(
2
→
3
)
⇒
s
(
2
→
3
)
=
4
4
=
1
m
Total distance moved
s
(
0
→
3
)
=
4
+
1
=
5
m
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.
In case we need to calculate displacement in the stated period.
Position of the body at
t
=
0
s
x
0
=
0
2
−
4
×
0
+
6
x
0
=
6
m
Position of the body at
t
=
3
s
x
3
=
3
2
−
4
×
3
+
6
x
3
=
9
−
12
+
6
x
3
=
3
m
Displacement in the given time interval
x
3
−
x
0
=
3
−
6
=
−
3
m
x
3
and
x
0
are displacements and therefore, vector quantities.
Distance is scalar.
Explanation: