Two particles A and B start from the same point
and move in the positive x-direction. In a time
interval of 2.00 s after they start, their velocities v
vary with time t as shown in the following figures.
What is the maximum separation between the
particles during this interval?
Answers
Answer:
1.25 m
Explanation:
Two particles A and B start from the same point and move in the positive x-direction. In a time interval of 2.00 s after they start, their velocities v
vary with time t as shown in the following figures.
What is the maximum separation between the particles during this interval?
Particles are with uniform speed for 1st sec
Then Different uniform speed for 2nd sec
so we will find Distance between both separately for 1st sec & 2nd sec
Particle A for t < 1
u = 2m/s a = 0
using S = ut + (1/2)at²
Distance S = 2t as no acceleration
Particle B for t < 1
a = ( 2 - 0)/1 = 2 m/s²
Distance S = 0t + (1/2)2t² = t²
Distance b/w 0 - 1 sec
ΔS = | 2t - t² |
dΔS/dt = |2 - 2t| = 0
=> t = 1
at t = 1 ΔS = | 2*1 - 1²| = |2 - 1|= 1 m
at t = 1 Particle A = 2 m , Particle B = 1 m
After t > 1
Particle A u = 1m/s a = 0
S = 2 + t ( 2 is distance covered till 1 sec)
Particle B
S = 1 + t² ( 1 is the distance covered till 2 sec)
ΔS = | 2 + t - 1 - t² |
=> ΔS = | 1 + t - t² |
dΔS/dt = -2t + 1
=> t = 1/2
this t = 1/2 is after 1st sec
Particle A = 2 + (1/2) = 2.5 m
Particle B = 1 + (1/2)² = 1 + 1/4 = 1.25 m
Maximum Separation at T = 1.5 sec
2.5 m - 1.25 m = 1.25 m
1.25 m
Answer:
1.25m
Explanation:
Two particles A and B start from the same point and move in the positive x-direction. In a time interval of 2.00 s after they start, their velocities v
vary with time t as shown in the following figures.
What is the maximum separation between the particles during this interval?
Particles are with uniform speed for 1st sec
Then Different uniform speed for 2nd sec
so we will find Distance between both separately for 1st sec & 2nd sec
Particle A for t < 1
u = 2m/s a = 0
using S = ut + (1/2)at²
Distance S = 2t as no acceleration
Particle B for t < 1
a = ( 2 - 0)/1 = 2 m/s²
Distance S = 0t + (1/2)2t² = t²
Distance b/w 0 - 1 sec
ΔS = | 2t - t² |
dΔS/dt = |2 - 2t| = 0
=> t = 1
at t = 1 ΔS = | 2*1 - 1²| = |2 - 1|= 1 m
at t = 1 Particle A = 2 m , Particle B = 1 m
After t > 1
Particle A u = 1m/s a = 0
S = 2 + t ( 2 is distance covered till 1 sec)
Particle B
S = 1 + t² ( 1 is the distance covered till 2 sec)
ΔS = | 2 + t - 1 - t² |
=> ΔS = | 1 + t - t² |
dΔS/dt = -2t + 1
=> t = 1/2
this t = 1/2 is after 1st sec
Particle A = 2 + (1/2) = 2.5 m
Particle B = 1 + (1/2)² = 1 + 1/4 = 1.25 m
Maximum Separation at T = 1.5 sec
2.5 m - 1.25 m = 1.25 m
1.25 mTwo particles A and B start from the same point and move in the positive x-direction. In a time interval of 2.00 s after they start, their velocities v
vary with time t as shown in the following figures.
What is the maximum separation between the particles during this interval?
Particles are with uniform speed for 1st sec
Then Different uniform speed for 2nd sec
so we will find Distance between both separately for 1st sec & 2nd sec
Particle A for t < 1
u = 2m/s a = 0
using S = ut + (1/2)at²
Distance S = 2t as no acceleration
Particle B for t < 1
a = ( 2 - 0)/1 = 2 m/s²
Distance S = 0t + (1/2)2t² = t²
Distance b/w 0 - 1 sec
ΔS = | 2t - t² |
dΔS/dt = |2 - 2t| = 0
=> t = 1
at t = 1 ΔS = | 2*1 - 1²| = |2 - 1|= 1 m
at t = 1 Particle A = 2 m , Particle B = 1 m
After t > 1
Particle A u = 1m/s a = 0
S = 2 + t ( 2 is distance covered till 1 sec)
Particle B
S = 1 + t² ( 1 is the distance covered till 2 sec)
ΔS = | 2 + t - 1 - t² |
=> ΔS = | 1 + t - t² |
dΔS/dt = -2t + 1
=> t = 1/2
this t = 1/2 is after 1st sec
Particle A = 2 + (1/2) = 2.5 m
Particle B = 1 + (1/2)² = 1 + 1/4 = 1.25 m
Maximum Separation at T = 1.5 sec
2.5 m - 1.25 m = 1.25 m
1.25 m