The First Problem

A car has to travel a distance s on a straight road. The car has maximum acceleration a and maximum deceleration b. It starts and ends at rest.

Show that if there is no speed limit, the time is given by

\displaystyle \sqrt{2s\cdot \frac{(a+b)}{ab}}.

Solution: We are going to use two pieces of information:

  • the area under the velocity-time graph is the distance travelled,
  • the slope of the velocity-time graph is the acceleration.

First, a rough sketch of the velocity-time graph:

graph3

The slope of the first line segment is a, while the slope of the second line segment is b. The area under the curve is given by s, the total distance. The base is t_1+t_2=T, the total time, and the height is the maximum speed.

Let v be the maximum speed reached. Using the area of a triangle, where T=t_1+t_2 is the total time:

\displaystyle s=\frac12 Tv \Rightarrow v=\frac{2s}{T}.

Now using the fact that the slope of the line segments, rise/run, is equal to the accelerations, we derive:

\displaystyle a=\frac{v}{t_1}\Rightarrow t_1=\frac{v}{a}, and

\displaystyle b=\frac{v}{t_2}\Rightarrow t_2=\frac{v}{b}.

Thus

\displaystyle T=t_1+t_2=\frac{v}{a}+\frac{v}{b}=\frac{va}{ab}+\frac{vb}{ab}=v\cdot \frac{(a+b)}{ab}.

Now, using v=\frac{2s}{T}:

\displaystyle T=\frac{2s}{T}\frac{(a+b)}{ab}

\displaystyle \underset{\times T}{\Rightarrow } T^2=2s\cdot \frac{(a+b)}{ab}.

Taking square roots (and noting T>0) completes the proof.

The Second Problem (1997)

A particle, moving in a straight line, accelerates uniformly from rest to a speed v. It continues at this constant speed for a time and then decelerates uniformly to rest, the magnitude of the deceleration being twice that of the acceleration. The distance traveled while accelerating is 6 m. The total distance traveled is 30 m and the total time taken is 6 s. 

i. Draw a velocity-time graph and hence, or otherwise, find v.

ii. Calculate the distance travelled at constant velocity.

Solution: 

i. First the velocity-time graph:

graph4

There are number of things we need to use:

  • the area under the velocity-time graph is the distance traveled
  • the slope of the velocity-time graph is the acceleration

The question gives the following:

  • the total distance is 30 m
  • the total time is 6 s
  • the distance traveled from t=0 to t=t_1 is 6 m.

In the first triangle:

\displaystyle a=\frac{v}{t_1}\Rightarrow t_1=\frac{v}{a}.

In the second triangle:

\displaystyle 2a=\frac{v}{t_2}\Rightarrow t_2=\frac{v}{2a}=\frac{1}{2}t_1,

and also t_1=2t_2. Define Let t:=t_2 so that t_1=2t. Therefore, using the fact that the total time is 6 s:

t_1+t_2+t_3=2t+t_2+t=6\Rightarrow t_2=6-3t.

Now, using the fact that the area under the first triangle is 6:

\frac{1}{2}t_1v=tv=6,

and the total area is 30:

6+vt_2+\frac12 tv=30

\Rightarrow 6+v(6-3t)+3=30

\Rightarrow 6+6v-3\underbrace{tv}_{=6}+3=30

\Rightarrow 6v=30-6+18-3=39\Rightarrow v=\frac{13}{2} m/s.

ii. Using tv=6 we have t=\frac{12}{13} so that the time spent at constant velocity is:

\displaystyle t_2=6-3t=6-\frac{36}{13}=\frac{42}{13},

and therefore the distance traveled is:

s_2=\frac{13}{2}\times \frac{42}{13}=21 \text{ m}.

Alternatively note that s_3=\frac12 s_1=3 and so s_2=30-s_1-s_2=21 m.

 

 

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