A particle moves along the s-axis. Use the given information to find the position of the particle. v(t)=8t3 ;
s(0) = 5.
yes i did mean
v(t) = 8 t^3 in the x direction.
i have a feeling some thing is wrong in your assumption of the question .. could u please recheck ?
I see Bob Pursley got the same answer I did.
To find the position of the particle, we need to integrate the given velocity function, v(t), with respect to time (t).
Given: v(t) = 8t^3
First, we integrate v(t) to obtain the position function, s(t):
s(t) = ∫v(t)dt
s(t) = ∫8t^3 dt
To integrate 8t^3, we can use the power rule of integration.
The power rule states that if we have a term of the form x^n, where n is any real number except -1, then the integral of x^n with respect to x is:
∫x^n dx = (1/(n+1)) * x^(n+1) + C
Applying the power rule to our integral:
s(t) = (1/4) * t^4 + C
Now, to determine the constant of integration, C, we use the initial condition provided:
s(0) = 5
Substituting t = 0 into the position function, we get:
(1/4) * (0^4) + C = 5
0 + C = 5
C = 5
Therefore, the position function of the particle is:
s(t) = (1/4) * t^4 + 5
This equation allows us to find the position of the particle at any given time, t, along the s-axis.
do you mean
v(t) = 8 t^3 ???
where v is velocity in x direction???
if so
s = s(0) + integral from 0 to t of 8 t^3 dt
= s(0) + 2 t^4
= 5 + 2 t^4