Question

substance of mass 5 kg is moving in y direction,force F=kt^2 is acting in the direction of motion,where k=15s^-2. The distance travelled by substance in the first 2 second_,and speed after 2 sec is_?
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Answer 1

Given:

Substance of mass 5 kg is moving in y direction,force F=kt^2 is acting in the direction of motion,where k=15s^-2.

To find:

• Distance in 2 sec ?
• Speed after 2 sec ?

Calculation:

$\therefore \: F = k {t}^{2}$

$\implies\: F = 15 {t}^{2}$

$\implies\: ma = 15 {t}^{2}$

$\implies\: 5a = 15 {t}^{2}$

$\implies\: a =3 {t}^{2}$

$\implies\: \dfrac{dv}{dt} =3 {t}^{2}$

$\implies\: dv =3 {t}^{2} \: dt$

$\implies\: \displaystyle \int dv =3 \int{t}^{2} \: dt$

$\implies\: \displaystyle \int_{0}^{v} dv =3 \int_{0}^{2}{t}^{2} \: dt$

$\implies\: \displaystyle v = {t}^{3} \bigg| _{0}^{2}$

$\boxed{ \implies\: \displaystyle v = 8 \: m {s}^{ - 1} }$

$\implies\: \displaystyle v = {t}^{3}$

$\implies\: \displaystyle \dfrac{dx}{dt} = {t}^{3}$

$\implies\: \displaystyle dx= {t}^{3} \: dt$

$\implies\: \displaystyle \int_{0}^{x} dx= \int_{0}^{2}{t}^{3} \: dt$

$\implies\: \displaystyle x = \frac{ {t}^{4} }{4} \bigg| _{0}^{2}$

$\implies\: \displaystyle x = \dfrac{16}{4}$

$\boxed{ \implies\: \displaystyle x = 4 \: m}$

Hope It Helps.