# Distance travelled given velocity at various moments of time [closed]

I am trying to solve a problem where I need to find the distance travelled at the end of the nth second and my input data contains only velocities at different time instances.

Say for instance I would like to find out the distance travelled at the end of the 5th second with the given data where each row corresponds to velocity at a different time instant

Time  Velocity
0       0
2      15
3      25
4      84
5     140


How should I approach this problem?

## closed as off-topic by jinawee, John Rennie, Brandon Enright, DavePhD, Qmechanic♦Apr 21 '14 at 9:41

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• Is the acceleration constant, or is it also changing? – LDC3 Apr 20 '14 at 16:58
• The acceleration is constant and was not mentioned – rgk Apr 20 '14 at 17:02
• Then it's simple. If you plot the time vs. velocity and calculate the slope, you get acceleration and using $d=\frac {1}{2}at^2$, you get distance. – LDC3 Apr 20 '14 at 17:05
• The acceleration is not constant. The average acceleration over the third second is $10 ms^{-2}$, the average acceleration over the fifth second is $56 ms^{-2}$ – DJohnM Apr 20 '14 at 17:34
• From plotting the data, $v(t)\sim2.2t^{2.55}$ which means $a(t)\sim t^{1.55}$; that is to say, the acceleration is definitely not constant. – Kyle Kanos Apr 20 '14 at 18:25

We wouldn't normally answer what is blatantly a homework problem, however I think there is an important principle here and it's instructive to go into detail. Any experimental physicist will tell you that when faced with experimental data the very first thing to do is DRAW A GRAPH:

The blue squares are your data and the red line is a fit (see below).

Having drawn the graph it is immediately obvious that the first three points lie on an approximately straight line and so do the second three points. To check this I did a linear regression of the first three and the last three points and the red lines are the result. The two straight lines are an excellent fit.

So the equation for the velocity as a function of time is:

\begin{align} v &= 8.21t - 0.36 \tag{t < 3} \\ &= 57.5t - 147 \tag{t > 3} \end{align}

where the numbers came from the linear regression (using Excel - other data analysis tools are available). The distance travelled can then be calculated by integrating this velocity time equation:

$$s = \int_0^3 (8.21t - 0.36) dt + \int_3^5 (57.5t - 147) dt$$

Since it's homework I'll leave you to do the legwork.