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Integral Calculus> The acceleration at t1 of a particle movi...

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The acceleration at t1 of a particle moving along the X-axisis given by a(t)=20 t^3 + 6. At time t=0 the velocity of the particle is 7.What is the position of the particle at the time t?

JOSEPH KIMUTAI , 5 Years ago

Grade 6

1 Answers

Vinod Ramakrishnan Eswaran

Last Activity: 5 Years ago

To find position we need to integrate acceleration two times to get distance(ie position)

a=20t³+6

v=∫a.dt+c ( c is a constant)

$v=\int (a.dt)+c=\int (20t^{3}+6)dt+c$

$v=\frac{20t^{4}}{4}+6t+c$

Given at t=0 v=7

$7=\frac{20(0)^{4}}{4}+6(0)+c$

which implies c=7

The new equation is

$v=\frac{20(t)^{4}}{4}+6(t)+7$

$v=5(t)^{4}+6(t)+7$

Integrating again wrt t we get

$s=\int v.dt +c_{1}=\int (5(t)^{4}+6(t)+7) +c_{1}$

where c1 is again constant of integration

$s= (5\frac{(t)^{5}}{5}+\frac{6(t)^{2}}{2}+7t) +c_{1}$

$s= ((t)^{5}+3(t)^{2}+7t) +c_{1}$

During t=0 usually object covers 0 distance therefore

$0= ((0)^{5}+3(0)^{2}+7* (0)) +c_{1}$

which implies c₁=0

Hence distance at time t is

$s= (t)^{5}+3(t)^{2}+7t$

Hope this answers your question

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