Question

A particle moves in a straight line so that its velocity at time

Answer 1

Answer:

s(2) = 7.75

Step-by-step explanation:

given the velocity v(t) = t^3

we can find the position s(t) by simply integrating v(t) and using the boundary conditions s(1)=2

$s(t) = \int {v(t)} \, dt\\ s(t) = \int {t^3} \, dt\\s(t) = \dfrac{t^4}{4}+c$

we know throught s(1) = 2, that at t=1, s =2. we can use this to find the value of the constant c.  

$s(1) = \dfrac{1^4}{4}+c\\4 = \dfrac{1^4}{4}+c\\c = 4-\dfrac{1}{4}\\c = \dfrac{15}{4} = 3.75$

Now we can use this value of t to formulate the position function s(t):

$s(t) = \dfrac{t^4}{4}+\dfrac{15}{4}$

this is the position at time t.

to find the position at t=2

$s(2) = \dfrac{2^4}{4}+\dfrac{15}{4}$

$s(2) = \dfrac{2^4}{4}+\dfrac{15}{4}$

$s(2) = \dfrac{31}{4} = 7.75$

the position of the particle at time, t =2 is s(2) = 7.75