Question

If velocity depends on time as :
=12t²
Find displacement in time interval between t= 0 and t = 2
options
1. 16
2. 24
3. 32
4. 48
please answer which option is correct ​

Answer 1

Given :

Velocity ,V = 12t²

To find :

Displacement in time interval between t = 0 and t = 2

Theory :

The rate of change of displacement of a particle with time is called velocity of the particle.

${\purple{\boxed{\large{\bold{Velocity=\frac{Distance}{Time\:interval}}}}}}$

In differential form:

$\sf\:Velocity,V=\dfrac{dx}{dt}$

Solution :

We have,to find displacement in interval [ 0 to 2 ]

$\sf\:V=12t^2$

$\sf\dfrac{dx}{dt}=12t^2$

$\sf\:dx=12t^2\:dt$

Now integrate both sides, within in the given interval

$\Big[x\Big]_{0}^{x} =\Big[12t^2\Big]_{0}^{2}dt$

We know ,that

$\sf\int\:x^n=\dfrac{x{}^{n+1}}{n+1}$

Then

$\sf\Big[x\Big]^x_0=\Big[\dfrac{12t{}^{2+1}}{2+1}\Big]^2_0$

$\sf\:x=\Big[\dfrac{12t^3}{3}\Big]^2_0$

$\sf\:x=\Big[4t^3\Big]^2_0$

$\sf\:x=4\times(2)^3-4\times0$

$\sf\:x=4\times8$

$\sf\:x=32m$

Therefore ,Displacement in time interval between t= 0 and t = 2 is 32m

$\rule{200}2$

More information about topic

1. Velocity is a vector quantity.
2. The velocity of an object can be positive, zero and negative.
3. SI unit: m/s
4. CGS unit : cm/s
5. Dimensions: $\sf\:[M^0LT{}^{-1}]$

Answer 2

Answer:

Given :

Velocity ,V = 12t²

To find :

Displacement in time interval between t = 0 and t = 2

Theory :

The rate of change of displacement of a particle with time is called velocity of the particle.

{\purple{\boxed{\large{\bold{Velocity=\frac{Distance}{Time\:interval}}}}}}

Velocity=

Timeinterval

Distance

In differential form:

\sf\:Velocity,V=\dfrac{dx}{dt}Velocity,V=

dt

dx

Solution :

We have,to find displacement in interval [ 0 to 2 ]

\sf\:V=12t^2V=12t

2

\sf\dfrac{dx}{dt}=12t^2

dt

dx

=12t

2

\sf\:dx=12t^2\:dtdx=12t

2

dt

Now integrate both sides, within in the given interval

\Big[x\Big]_{0}^{x} =\Big[12t^2\Big]_{0}^{2}dt[x]

0

x

=[12t

2

]

0

2

dt

We know ,that

\sf\int\:x^n=\dfrac{x{}^{n+1}}{n+1}∫x

n

=

n+1

x

n+1

Then

\sf\Big[x\Big]^x_0=\Big[\dfrac{12t{}^{2+1}}{2+1}\Big]^2_0[x]

0

x

=[

2+1

12t

2+1

]

0

2

\sf\:x=\Big[\dfrac{12t^3}{3}\Big]^2_0x=[

3

12t

3

]

0

2

\sf\:x=\Big[4t^3\Big]^2_0x=[4t

3

]

0

2

\sf\:x=4\times(2)^3-4\times0x=4×(2)

3

−4×0

\sf\:x=4\times8x=4×8

\sf\:x=32mx=32m

Therefore ,Displacement in time interval between t= 0 and t = 2 is 32m

$$\rule{200}2$$

Explanation:

hope it helped you friend please mark me as brainliest and follow me I need more thanks please