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Motion in a Plane

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Question : 21
Total: 32
A particle starts from the origin at t=0s with a velocity of 10.0
^
j
 m/s and moves in the x-y plane with a constant acceleration of (8.0
^
i
+2.0
^
j
)msāˆ’2 (a) At what time is the x-coordinate of the particle 16 m? What is the y-coordinate of the particle at that time ? (b) What is the speed of the particle at the time?
Solution:  
Here,
ā†’
u
=10.0
^
j
msāˆ’1 at t=0
ā†’
a
=
d
ā†’
v
dt
=(8.0
^
i
+2.0
^
j
)msāˆ’2 So ,d
ā†’
v
=(8.0
^
i
+2.0
^
j
)dt
Integrating it within the limits of motion i.e. as time changes from 0 to t, velocity changes from u to v, we have
ā†’
v
āˆ’
ā†’
u
=(8.0
^
i
+2.0
^
j
)t or
ā†’
v
=
ā†’
u
+8.0t
^
i
+2.0t
^
j
As
ā†’
v
=
d
ā†’
r
dt
 or d
ā†’
r
=
ā†’
v
dt So, d
ā†’
r
=(
ā†’
u
+8.0t
^
i
+2.0t
^
j
)dt
Integrating it within the conditions of motion i.e. as time changes from 0 to t, displacement is from 0 to r, we have
ā†’
r
=
ā†’
u
t+
1
2
Ɨ8.0t2
^
i
+
1
2
Ɨ2.0t2
^
j

or x
^
i
+y
^
j
 =10t
^
j
+4.0t2
^
i
+t2
^
j
 =4.0t2
^
i
+(10t+t2)j
Here, we have, x=4.0t2 and y=10t+t2āˆ“t=(
x
4
)
1
2

(a) ā€Œ At x=16m ; t=(
16
4
)
1
2
=2s, y=10Ɨ2+22=24m
(b) Velocity of the particle at time t is
ā†’
v
=10
^
j
+8.0t
^
i
+2.0t
^
j

When t=2s, then
ā†’
v
=10
^
j
+8.0Ɨ2
^
i
+2.0Ɨ2
^
j
 =16
^
i
+14
^
j

Speed =|
ā†’
v
|=āˆš162+142=21.26msāˆ’1
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