A body is moving along the +ve x- axis with uniform acceleration of -4ms-2 . it's velocity at x=0 is 10ms-1 . the time taken by the body to reach a point at x=12m is ?
Answers
Answer:
Explanation:
Given data
The magnitude of the acceleration is:
a
=
−
4
m
⋅
s
−
2
The velocity at
x
1
=
0
m
is:
u
=
10
m
⋅
s
−
1
Use the equation to calculate the time required to reach the point at
x
2
=
12
m
x
2
−
x
1
=
u
t
+
1
2
a
t
2
Here, t is the required time.
Substitute all the values in the above equation.
x
2
−
x
1
=
u
t
+
1
2
a
t
2
12
−
0
=
(
10
)
t
+
1
2
(
−
4
)
t
2
12
=
10
t
−
2
t
2
2
t
2
−
10
t
+
12
=
0
t
2
−
5
t
+
6
=
0
t
2
−
3
t
−
2
t
+
6
=
0
Further, solve the above equation.
t
(
t
−
3
)
−
2
(
t
−
3
)
=
0
(
t
−
3
)
(
t
−
2
)
=
0
(
t
−
3
=
0
)
o
r
(
t
−
2
=
0
)
(
t
=
3
s
)
o
r
(
t
=
2
s
)
The object first reaches the point
x
2
in
2
s
in its forward journey but as we can observe that the acceleration is negative so the object will again cross the point
x
2
during its backward journey at
3
s
hope this helpsyou and others please mark me as brainliest
Answer:
(2s, 3s)
Explanation:
a = (-4) m/s^2, u = 10 m/s, s = 12m
s = ut + 1/2 at^2
12 = 10t + 1/2 (-4)t^2
12 = 10t - 2t^2
2t^2 - 10t + 12 = 0
If you simply the equation you will get
t = 2s and 3s