Question

hey all

A particle starts moving rectilinearly at t = 0 such that it's Velocity v changes with time according to the relation
v = t^2 - t. Find time interval for which particle retardes.


Need proper SOLUTION.​

Answer 1

Answer- The above question is from the chapter 'Kinematics'.

Some basic formulae:

1) Velocity = $\frac{dx}{dt}$

2) Acceleration = $\frac{dv}{dt}$

3) Acceleration (when x is given) = v $\frac{dv}{dx}$

Given question: A particle starts moving rectilinear at t = 0 such that its Velocity v changes with time according to the r*elation v = t² - t. Find time interval for which particle decelerates.

Answer: v = t² - t

a = $\frac{dv}{dt}$ t² - t

a = 2t - 1

Now, velocity and acceleration should be in opposite direction.

Case - 1

At t = 0 seconds,

a = 2(0) - 1

a = -1 m/s² which is negative.

a should be less than zero, then only it can decelerate.

2t - 1 < 0

2t < 1

t < 0.5 s

⇒ 0 < t < 0.5  --- (1)

Case - 2

a should also be greater than zero, then only it can get back to its original position and deceleration occurs.

2t - 1 > 0

2t > 1

t > 0.5 s

At t = 1 seconds,

a = 2(1) - 0

a = 1 m/s² which is positive.

⇒ 0.5 < t < 1   --- (2)

From equations 1 and 2, we get

0.5 < t < 1 is the required time interval.

Answer 2

Answer:

1/2<t<1

Step-by-step explanation:

V= t^2-t

** a = dv/dt = 2t-1

* motion is considered as retards when Vand a are in opposite direction

CASE - 1

* if v>0 then a<0

* but t^2-t >0,t>1

* and a>0 for t>1

* so not possible

** CASE - 2

* v<0,a>0

* t^2-t<0,2t-1. 0

* t belongs to (0,1),t>1/2

********1/2<t<1**********