a question about kinematics of a particle

[losheungwai]

a car decelerates on a ramp with an initial velocity of 100 km/h and final velocity of 0.

how to find the time at which the car stops and the distance traveled on the ramp given that dv/dt varies linearly from 0 to -3 m/s^2 during evry time interval of 1s????

 

[halls]

ok, here it goes what i understood:

A car is climbing a ramp, with an initial velocity of 100 km/h. At every second the car suffers a deceleration which goes from 0 to -3 m/s^2 during the whole second, linearly. So the acceleration the car would suffer would be:

second 1.0 : 0 m/s^2
second 1.1 : -0.3 m/s^2
second 1.2 : -0.6 m/s^2
.
.
.
second 1.9 : -2.7 m/s^2
second 2.0 : 0 m/s^2
second 2.1 : -0.3 m/s^2
second 2.2 : -0.6 m/s^2
.
.
.
and so on...

If this is right, then here it is my solution:

First of all we have to see the whole time needed as if it were N time intervals of one whole second plus an interval of \[t_x\] seconds, this is:

\[t_{total}=n + t_x\], where n is an integer value.

So, let's se what happens in a 1 second interval:

As it is said, acceleration will vary from 0 to -3 m/s^2 linearly during the whole second, this is:

\[a = -3t = \frac{dv}{dt}\], so \[dv = -3t\, dt\]

Then:

\[\int_{v_o}^{v} dv = \int_{t_o}^{t} -3t\, dt\]

So developing the equation: \[v = v_o + \frac{3}{2}(t_o^2 - t^2)\]

and as in every interval, we take the time reference as a new one, \[t_o=0\], so:

\[v = v_o - \frac{3}{2}t^2\]

Now, as after every interval, our time \[t=1\] second, the velocity formula for the end of every interval will be: \[v = v_o - 3/2\] (remember we are working in meters and seconds). So, if we take a look at the 1 second intervals from the begining, we'll se a pattern:

\[v_o = 100 km/h = 27.7778 m/s\]
\[v_1 = v_o - \frac{3}{2} = 26.2778 m/s\]
\[v_2 = v_1 - \frac{3}{2} = 24.7778 m/s\]
...
\[v_n = v_o - \frac{3}{2}n\]

So how many 1-second intervals will we be having? Easy, \[v_n = 0 = v_o - \frac{3}{2}n\].
\[n = \frac{v_o . 2}{3} = 18.518 seconds\]

But as intervals have to be discrete (we can't have half interval, just one whole interval) we take the lowest integer: 18.

So now we know our total time will be 18 seconds plus something else which is less than a second... Now it's time to calculate that one. For that we take again the formula we calculated for an interval, but this time, the end time won't be 1 second, but an unknown variable. And this time, \[v_o\] won't be 100 km/h either, but the speed at which the car arrived at the last interval:
\[v_o = 27.77778 - \frac{3}{2}.18 = 0.77778 m/s\]

And now we also know the final velocity for the car, for what we know all variables needed, taking again the formula:

\[v = v_o + \frac{3}{2}(t_o^2 - t^2)\]

and we substitute the values:

\[ 0 = 0.77778 + \frac{3}{2}(0 - t^2)\]

If we operate that, we will finally get:

\[t = \sqrt{ \frac{2 \cdot 0.77778}{3}} = 0.72 seconds\]

Then finally, our total time, this is, the time the car needs to stop moving, will be:

\[T = 18 + 0.72 = 18.72 seconds\]

Hope it's clear and helps