Question

Given the acceleration, initial velocity, and initial position of a body moving along a coordinate line at time t, find the body's position at time t.
a=16, v(0)= -12, s(0)= 13

Answer 1

first find t by using a=v-u/t

Answer 2

u=?

Answer 3

did the body start from rest

Answer 4

Is initial position a? it just says that and that it is moving

Answer 5

why dont u try u =0

Answer 6

the answer is s= 8t^2 - 12t+13 I just don't know how to get there

Answer 7

just integrate starting with the acceleration:
\large \int 12 dt= ???? this will give you the velocity.

use the intial velocity to find the constant of integration.

then integrate this velocity to find the position function. use the initial position to determine the constant of integration.

Answer 8

@ct3324432089 what math level are you in?

Answer 9

Well, you know that acceleration is the derivatie of velocity with respect to time, so you have \(\int dt\)= t + C where youre solving for C.

however, you didn't give a complete initial condition for a(t) = 0
is it a(t) = 16 or a(0) = 16?? because you get different answers for both.

Answer 10

my problem just says a=16. I think this is where i'm getting confused. I've got the problem and the answer, but not the means

Answer 11

then your answer is 16t for velocity. \(\int 16dt = 16t\)

Answer 12

do I plug that back into the a=v-(u/t)? I thought I had to find the antiderivative?

Answer 13

the formula works for every t.
we have three components in the formula:

a) the initial position. it was 13, more time does not change the offset
b) the velocity, it does not decrease so it will add -12m EVERY t (-12 x t)
c) the distance from acceleration,,,,,,,

Answer 14

constant acceleration results in additional velocity:
t=0: 16m/s
1: 32m/s
2: 48m/s
3: 64m/s
4: 80m/s
...

Answer 15

yeah you need the antiderivative to get back the distance from a(t)

Answer 16

gotcha, thanks a ton guys and ladies

Answer 17

integrate