Question

q2 and 3 both. pls answer my question.i will mark as brainliest if correct and report if wrong​

Answer 1

ques=2

v=2x²

as a=dv/dx

a=d(2x²)/dx

=4x

at x= 2m  

a=4*2

8m/s2

question 3

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