the distance travelled by a particle moving along a straight line is given by x is equal to 183 + 15 30 square metre find first the initial velocity of the particle second the velocity at the end of second and third the acceleration of the particle
Answers
When acceleration is opposite to the direction of velocity (I assume this is what you mean by negative acceleration), the object will certainly reverse its direction of motion at some instant. If you want to find the distance (not displacement) travelled in some time interval, you have to check whether the instant of zero velocity falls in that interval. It does in your case at t=2.5 s as Aditya pointed out in his answer and I reproduce it here.
v=u+at=>0=12.5–5t=>t=2.5s
v
=
u
+
a
t
=>
0
=
12.5
–
5
t
=>
t
=
2.5
s
This means distance travelled in 3rd second = distance travelled between 2 s to 2.5 s in forward direction + distance travelled between 2.5 s and 3 s in backward direction.
Now the distance travelled im 2.5 s is
(12.5).(2.5)+(1/2).(−5).(2.5)2=15.625m
(
12.5
)
.
(
2.5
)
+
(
1
/
2
)
.
(
−
5
)
.
(
2.5
)
2
=
15.625
m
And the distance travelled in 2 s is
(12.5).2+1/2.(−5).22=15m
(
12.5
)
.2
+
1
/
2.
(
−
5
)
.2
2
=
15
m
Thus the distance travelled between 2 s to 2.5 s is 15.625–15=0.625m
15.625
–
15
=
0.625
m
The same distance will be travelled between 2.5 s to 3 s.
Hence the total distance travelled in 3rd second is (0.625).2=1.25m