Simple Dynamics

My granddaughter wanted help on her Physics GCSE question. Great as I know she will not continue with this subject next year so give me an opportunity to help. This relatively simple can be used to introduce derivation of basic dynamic equations and the concepts of integration without any previous knowledge of calculus. Further if calculus is understood can be further developed to show how easy the basic dynamic laws can be derived using calculus.

Question

A car is driven at a constant speed past a speed camera, which recorded two images of the car 0.70 s apart. The car travelled 14 m between the two images. The maximum deceleration of the car is 6.25 m/sec2.

Calculate the minimum braking distance for the car at the speed it passed the speed camera.

Answer

Speed = Distance/Time or v=d/t. where v is current speed. (I am treating velocity as a scalar in this example)

So v = 14/0.70 = 20 m/s.

So far this is simple and my granddaughter had worked this out herself.

I recalled (from 50+ years ago) the equation for constant acceleration v2 = u2 + 2ad where v is final velocity, u is initial velocity and a is acceleration.

In this problem v = 0 and a = -6.25 (- because it is deceleration).

v2 = u2 + 2ad

0 = (20)2 - 2(6.25)d

d = 400/12.5 = 32 m

Granddaughter pleased but how do I know if they have taught this equation, and even if they had did my granddaughter understand it. So how to answer without using the formula.

Figure 1 show the speed - time graph when velocity is constant, and how

distance = speed / time

Figure 1

acceleration = (change in speed) / time

Although not stated it is assumed that acceleration is constant which means the change of speed is the same for equal amount of time . In this problem the car reduces speed of 6,25 m/s for each second. Therefore the graph in Figure 2 is a straight line. In this problem we are dealing with deceleration, but we treat this as negative acceleration.

The initial speed is 20 m/s as shown above. So time to come to a stop (speed = 0) is

20 / 6.25 = 3.2 s.

The more interesting part of the problem is how to determine the distance travelled in coming to a stop.

First approach which I used to explain to my granddaughter makes the reasonable assumption that if we calculate the average speed and multiply by time we shall derive the distance.

aver. speed = ( 20 - 0 ) / 2 = 10 m/s

distance = 10 x 3.2 = 32 m.

So answer is 32m

Figure 2

Derive standard equations

Consider a body has initial speed of u, final speed of v,and uniform acceleration of f.

As shown above the distance traveled, s, is the sum of the distance travelled due to initial speed of u plus distance travelled because of acceleration f.

This is the area under the slope of the graph shown in Figure 3.

Speed = distance/time

v = s/t ...... (1)

acc. = (change of speed) / time

f = (v - u) / t

v = u + ft ...... (2)

(remember f is constant)

s = ut + ½ft² ...... (3)

From (2)

ft = (v-u)

t = (v-u)/f

Substitute in (3)

s = u(v-u)/f + ½f(v-u)²/f²

s = u(v-u)/f +½(v-u)²/f

Multiply both sides by 2f

2fs = 2uv - 2u² + v² -2uv + u²

2fs = -u² + v²

v² = u² + 2fs ...... (4)

Figure 3

Proof that we can use average speed x time

This is more of an explanation than a proof.

Figure 3 shows the chart from a spread sheet data shown in Figure 4. The time interval is 0.2 secs chosen as this is a factor of the time calculated to stop, i.e. 3.2 secs. The Distance - Max column use the max speed at the start of each 0.2 sec., the Distance Min uses the minimum speed, so creating maximum and minimum values for the distance travelled. Figure 4 shows the total distance travelled for both these situations and also taking the mid value between the two. The total distance travelled lies between 34 and 30 m.

Figure 3

Figure 4

Another spreadsheet was created, this time dividing the time into 80 parts, which is an interval of 0.04 secs. The maximum and minimum distances travelled is now 32.4 and 31.6 m. The shape of the graph also approaches that of a triangle. In both cases if the mid value of the time interval is used the distance travelled is 32m.

Figure 5

As the number of intervals is increased the shape becomes increasingly that of a triangle and the values for both maximum and minimum becomes ever closer to 32. So the distance travelled is that of the area of the graph which in this case is the area of the triangle i.e. ½vt = 10 x 3.2 = 32m

Determining equations using calculus

The reason for the "proof" above is because a very similar approach is used in calculus which analyses the rate of change of the slope of graphs, differentiation, and the calculation of areas beneath the graph, integration. Most equations and their plots are not straight lines, and so the actual values are not equal to the mid value as it was above.

I find calculus particularly enjoyable and I have a section on this site. CALCULUS

The following shows how quickly the equations can be derived using calculus. I have not explained this but included it to show how powerful it can be. It will be obvious to any one with a knowledge of basic calculus.

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