A Particle moves along a straight line so that it's velocity, v ms^-1 at time t seconds is given by v=6e^(3t) +4. When t = 0, the displacement, s, of the particle is 7 metres. Find an expression for s in terms of t.
I know im supposed to use integration but im a bit confused on how, can someone please explain how to do it. Thanks!

Question

anonymous · Answer

$$\int 6e^{3t}+4  dt$$

anonymous · Answer

2 e^3t+ 4t +5

cathyangs · Answer

go imranmeah91! :D I agree with what little knowledge I know. 

did you type this up? XD or is particle supposed to be capitalized?

ash2326 · Answer

we are given
v=6e^(3t)+4
v=dx/dt ,x=displacement
dx/dt=6e^(3t)+4
for getting a relation of x , we have to integrate the above eq.
on integration we get
x=6e^(3t)/3+4t+c   c=constant of integration
x=2e^3t+4t+c
it's given when t=0 x=7
put t=0
7=2+c
so c=5
x=2e^3t+4t+5

anonymous · Answer

Is the equation for integrating it not x^n+1 /n +1, so where is the plus one part?

ash2326 · Answer

but here we don't have any t term, and integral of constant like 4 is 4t
or u can say 4t^0---->4t^(0+1)/(0+1)=4t

anonymous · Answer

How come 4 becomes 4t?

anonymous · Answer

4 = 4 t^0

ash2326 · Answer

lau 4 can also be written as 4t^(0)

integral of 4t^0=4t^(0+1)/(0+1)
i.e 4t