Question

The velocity of the particle at any time t is given by v=2t(3-t)ms^1. At what time is its velocity maximum​

Answer 1

QUESTION:

The velocity of the particle at any time t is given by v=2t(3-t)ms^1. At what time is its velocity maximum.

ANSWER:

time at which velocity is maximum = 1.5 seconds

Explanation:

given that,

The velocity of the particle at any time t is given by v=2t(3-t)ms^1.

so,

here,

v is the function of t

and

v = 2t(3 - t)

v = 6t - 2t²

to find,

At what time is its velocity maximum

we know that,

for the maximum or minimum value,

dv/dt = 0

d(6t - 2t²)/dt = 0

after differentiation of dv/dt we get

6 - 4t = 0

-4t = -6

t = -6/-4

t = 3/2

t = 1.5 s

so,

___________________

time at which velocity is maximum

= 1.5 seconds

____________________

for the maximum velocity

putting the value of t as 1.5

v = 2t(3 - t)

putting the value of t

2(1.5)(3 - 1.5)

3(1.5)

= 4.5

so,

maximum velocity = 4.5 m/s² at t = 1.5 s

Answer 2

Answer:

V = 2t(3 - t)ms^2

= 6t -2t^2

We know that

a = dv/dt

= 6 - 4t

Velocity is maximum when acceleration of the body is zero

So 6 - 4t = 0

4t = -6

t= -6/4

t = -3/2

t = 1.5 sec

So when time = 1.5 sec velocity will be maximum.