Quite a simple question but I'm having issues -
I have the speed of an object (say 5m/s) and the time it takes the object to come to a complete stop (say 2s). I want the object to stop exactly on a given position (T). At what distance from T will I need to start decelerating?
The object has no friction.
On
Assuming a constant deceleration $ a $, and given the time it takes to fully stop; $ \Delta t =2 s$ and initial velocity $ v_{o} = 2m/s $ we can integrate the acceleration with respect to time once to find that the velocity at time $t$ is $v=at + v_{o}$ or in our specific case $v(t)=at+2$.
Since we know that after $\Delta t = 2s$ we come to a complete stop then we can use the above equation to find that: $ 0=a\cdot2s + 2 $ or equivalently $ 0 = a\cdot \Delta t + v_{0} $ (since the final velocity is zero) which gives the expression for the acceleration (or in this case deceleration): $ a = -\frac{v_{o}}{\Delta t} $ and for the specific case $ a=-1m/s^{2} $.
Now we integrate the velocity with respect to time to find the distance at a given time: $ s(t)=\frac{1}{2}at^{2} + v_{o}t + s_{o} $.
We set the initial position ($ s_{o} $) equal to zero such that: $ s(t)=\frac{1}{2}at^{2} + v_{o}t $, or in this specific case; $ s(t)=-(1)\frac{1}{2}t^{2} + 2t $.
Now all that is needed is to insert the time it takes to stop ($\Delta t$) into the above equation and we have the distance the object (the car) takes to fully stop: $ T=s(\Delta t)=\frac{1}{2}(-\frac{v_{o}}{\Delta t})\Delta t^{2} + v_{o}\Delta t $.
Inserting the values for the special case we get that: $ T=\frac{1}{2}(-1m/s^2)(2s)^{2} + (2m/s)(2s)=1m $
We can simplify the general expression above a bit and get that: $ T= \frac{1}{2}v_{o}\Delta t $.
I hope this helps.
The distance is the time multiplied by the average speed,
$$2\frac{5+0}2=5\text{m}.$$
(We assume uniform acceleration, so that the average speed is the mean of the initial and final speed.)