Question

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?
​

Answer 1

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 2

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