Question

The acceleration of a particle is directly proportional to the square of the time t. When t 0, the particle is at x 24 m. Knowing that at t 6 s, x 96 m and v 18 m/s, express x and v in terms of t

Answer 1

Given:

The acceleration of a particle is directly proportional to the square of the time t. When t=0, the particle is at x=24 m. Knowing that at t=6 s, x = 96 m and v= 18 m/s,

To find:

To express x and v in term t ?

Calculation:

$\rm \: a \propto {t}^{2}$

$\rm \implies \: a = k{t}^{2}$

$\implies \: \displaystyle \rm \int_{0}^{18} dv = k \int_{0}^{6}{t}^{2} \: dt$

$\implies \: \displaystyle \rm 18 = k \times \dfrac{216}{3}$

$\implies \: \displaystyle \rm 18 = k \times 72$

$\implies \: \displaystyle \rm k = \dfrac{1}{4}$

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

$\implies \: \displaystyle \rm v = \dfrac{k {t}^{3} }{3} + c$

$\implies \: \displaystyle \rm v = \dfrac{ {t}^{3} }{12} + c$

At t = 6, value of v = 18, hence c = 0.

$\boxed{ \implies \: \displaystyle \rm v = \dfrac{ {t}^{3} }{12} }$

$\implies \: \displaystyle \rm dx= \dfrac{ {t}^{3} }{12} \: dt$

$\implies \: \displaystyle \rm \int dx= \int \dfrac{ {t}^{3} }{12} \: dt$

$\implies \: \displaystyle \rm x= \dfrac{ {t}^{4} }{48} + c$

When t = 0, x = 24 , hence c = 24.

$\boxed{ \implies \: \displaystyle \rm x= \dfrac{ {t}^{4} }{48} + 24}$

Hope It Helps.