Question

A particle is moving with acceleration a(t)=30t+8. its position at time t=0 is s(0)=14 and its velocity at time t=0 is v(0)=13. What is its position at time t=4?

Answer 1

Answer:

s(4)=450

Explanation:

Given Data

a(t)=30t+8

at t=0 s(0)=14

at t=0 v(0)=13

s(4)=?

Solution

we know v is derivative of a

and s is derivative of v

a(t)=30t+8

derivative it we get

$v(t)=15t^{2}+8t+c\\ put t=0\\v(0)=15(0)^{2}+ 8(0)+c\\v(0)=c\\c=13$

so

$v(t)=15t^{2}+8t+13$

derivative it we get

$s(t)=\frac{15}{3}t^{3}+4t^{2}+13t+c\\ t=0\\s(0)=\frac{15}{3}(0)^{3}+4(0)^{2}+13(0)+c\\s(0)=c\\c=14$

so

$s(t)=\frac{15}{3}t^{3}+4t^{2}+13t+14\\t=4\\s(4)=\frac{15}{3}(4)^{3}+4(4)^{2}+13(4)+14\\s(4)=450$