1
Question
A particle starts from the origin at 𝑡=0 𝑠 with a velocity of 10.0^j m/s and moves in the x-y plane with constant acceleration of 8.0^i+2.0^j ms−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 ?
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 ?
Open in App
Solution
A) Step 1 : - Find →v(t) from relation →a(t)=d→v(t)dt
∫d→v(t)=∫→a(t)dt
Given →a(t)=8^i+2^j
∫v10^jd→v(t)=∫t0(8^i+2^j)dt
→v(t)−10^j=8t^i+2t^j
→v(t)=8t^i+2t^j+10^j
→v(t)=8t^i+(2t+10)^j.......(1)
Step 2 : - Find →r(t) from relation →v(t)=d→r(t)dt
As d→r(t)=→v(t)dt
∫r0d→r(t)=∫t0→v(t)dt
→r(t)=∫t0[8t^i+(2t+10)^j]dt
→r(t)=8t22^i+(2t22+10t)^j
We can write →r(t)=x(t)^i+y(t)^j=4t2^i+(t2+10)^j
From comparsion: -
x(t)=4t2
Putting x(t)=16 m
16=4t2
So, t = 2 s is the time when particle is at x = 16 m.
At t = 2s the y-coordinate of particle,
y(t)=t2+10=(2)2+10=14 m
Final Answer: The particle at x = 16 m is at t = 2s and at t = 2s the y-coordinate of particle is 14 m.
B) Step 1 : Find the magnitude of velocity at t=2 sec.
From equation (1)
→v(t)=8t^i+(2t+10)^j
Velocity at t = 2 s,
→vt=2s=8(2)^i+(2(2)+10)^j
→vt=2s=16^i+14^j
speed of particle : -
|→v|t=2s=√(16)2+(14)2
=√256+196
=√452
=21.26 m/s
Final Answer: Speed of the particle is 21.26 m/s.
∫d→v(t)=∫→a(t)dt
Given →a(t)=8^i+2^j
∫v10^jd→v(t)=∫t0(8^i+2^j)dt
→v(t)−10^j=8t^i+2t^j
→v(t)=8t^i+2t^j+10^j
→v(t)=8t^i+(2t+10)^j.......(1)
Step 2 : - Find →r(t) from relation →v(t)=d→r(t)dt
As d→r(t)=→v(t)dt
∫r0d→r(t)=∫t0→v(t)dt
→r(t)=∫t0[8t^i+(2t+10)^j]dt
→r(t)=8t22^i+(2t22+10t)^j
We can write →r(t)=x(t)^i+y(t)^j=4t2^i+(t2+10)^j
From comparsion: -
x(t)=4t2
Putting x(t)=16 m
16=4t2
So, t = 2 s is the time when particle is at x = 16 m.
At t = 2s the y-coordinate of particle,
y(t)=t2+10=(2)2+10=14 m
Final Answer: The particle at x = 16 m is at t = 2s and at t = 2s the y-coordinate of particle is 14 m.
B) Step 1 : Find the magnitude of velocity at t=2 sec.
From equation (1)
→v(t)=8t^i+(2t+10)^j
Velocity at t = 2 s,
→vt=2s=8(2)^i+(2(2)+10)^j
→vt=2s=16^i+14^j
speed of particle : -
|→v|t=2s=√(16)2+(14)2
=√256+196
=√452
=21.26 m/s
Final Answer: Speed of the particle is 21.26 m/s.
Suggest Corrections
0
and moves in the x-y plane
with a constant acceleration of
.