Question

A particle initially at rest moves along x-axis from origin. Its acceleration varies with time as a=(6t+5)m/s. The distance covered in first 2 sec is (in multiples of 6m)

Answer 1

Given:

Acceleration of particle,a=(6t+5)m/s

To Find:

The distance covered in first 2 sec

Solution:

It is given that,

$\longrightarrow\rm{a=6t+5}$

On integrating both the sides, we get

$\longrightarrow\rm{\displaystyle\int adt=\displaystyle\int(6t+5)dt}$

$\longrightarrow\rm{v=\dfrac{6t^{2}}{2} +5t}$

$\longrightarrow\rm{v=3t^{2}+5t}$

where, v is velocity

Now, on integrating both the sides again, we get

$\longrightarrow\rm{\displaystyle\int vdt=\displaystyle\int(3t^{2}+5t)dt}$

$\longrightarrow\rm{x=\dfrac{3t^{3}}{3}+\dfrac{5t^{2}}{2}}$

$\longrightarrow\rm{x=t^{3}+\dfrac{5t^{2}}{2}}$

where, x is the displacement of particle

On putting t=2 ,we get

$\longrightarrow\rm{x=(2)^{3}+\dfrac{5(2)^{2}}{2}}$

$\longrightarrow\rm{x=8+\dfrac{5(4)}{2}}$

$\longrightarrow\rm{x=8+\dfrac{20}{2}}$

$\longrightarrow\rm{x=8+10}$

$\longrightarrow\rm\green{x=18\ m}$

Answer 2

Answer

Equation of acceleration w.r.t time $\longrightarrow$

a = 6t + 5 sec.

Formula used -

$\int \: adt$ = Velocity

$\int \: vdt$ = Distance

Solution -

$\int adt = \int 6t + 5$

Integrating by using$\int {x}^{n} dx = \frac{ {x}^{n + 1} }{n + 1}$

v = $3 {t}^{2} + 5t$

Integrating both side -

$\int vdt = \int 3 {t}^{2} + 5t$

x = ${t}^{3} + {\frac{5t}{2} }^{2}$

At t = 2 sec

x = ${2}^{3} + \frac{ {5 \times 2}^{2} }{2}$

x = 8 + 10

x = 18 m

Thanks