Question

The performance of two different cars, car 1and car 2, was measured on a long horizontal test track. Car 1started from rest and ran with constant acceleration until it was halfway down the track and then stopped accelerating, continuing to run at the attained speed to the end of the track. Car 2started from rest and ran with a constant acceleration for the entire distance. It was observed that both cars covered the test distance in the same amount of time. (a) What was the ratio of the average speed of carto that of car 2? (b) What was the ratio of the initial acceleration of car 1to that of car 2? (c) What was the ratio of the final speed of car 1to that of car 2?

Short answer

a) The ratio of the average speed of car 1 to that of car 2 is 1 .

b) The ratio of the initial acceleration of car 1 and car 2 is $\frac{9}{8}$.

c) The ratio of the final speed of car 1 to that of car 2 is$\frac{3}{4}$.

Step-by-step solution

Definition of speed.

• Speed is the pace at which an object moves along a path in terms of time, whereas velocity is the rate and direction of movement.
• In other words, while speed is a scalar value, velocity is a vector.

The ratio of the average speed of car 1 to that of car 2 .

(a)

The average speed of the first car,$V_{avg.1}=\frac{∆x_1}{∆t_1}$

where the average speed of second car, $V_{avg.2}=\frac{∆x_2}{∆t_2}$

However, because the vehicles' tracks are the same length (i.e.$∆x_1=∆x_2$) and they travelled the same distance in the same amount of time (i.e.$∆t_1=∆t_2$), the ratio of car 1's average speed to that of car 2 is

$$\begin{gathered} \frac{V_{avg.1}}{V_{avg.2}}=\frac{∆x_1/∆t_1}{∆x_2/∆t_2} \\ =\frac{∆x/∆t}{∆x/∆t} \\ =1 \end{gathered}$$

The ratio of initial acceleration of car 1 to that of car 2 .

(b)

Because Car 1 begins its motion with a net force acting on it (remember,${\overrightarrow{F}}_{net}=m\overrightarrow{a}$), and the net force on it eventually becomes zero, we must split its motion into two stages. The automobile begins at rest, then undergoes a constant net force (i.e. constant acceleration) resulting in a final speed,

$$\begin{gathered} v_{1f}=v_{1i}+\frac{F_{net.1}}{m}∆t \\ =0+a_1\left(t-0\right) \\ =a_{1}t..........\left(1\right) \end{gathered}$$

When the force on the vehicle is zero (i.e. the acceleration is zero), the car will begin to move at the speed it was at in the previous phase (which is$v_{1}f$), and it will do so until it reaches the end of the track.

Now split the track into two sections, the first of which begins at $x=0$and ends at $x=\frac{d}{2}$and the second of which begins at $x=\frac{d}{2}$and ends at $x=d$.

The automobile will accelerate for the first half of the complete distance, and moving this distance will take time $∆t=t\text{'}$.

Therefore,

localid="1656675751124" $$\begin{gathered} ∆x=v_{1i}∆t+\frac{1}{2}\frac{F_{net}}{m}∆t^2........\left(2\right) \\ \frac{d}{2}=\frac{1}{2}{a}_1{t}^{\text{'}2}...........\left(3\right) \end{gathered}$$

This speed will be the same as the automobile's speed in the second half of the distance $x=\frac{d}{2}$and the car will require a time to reach the distance $∆t=t\text{'}\text{'}$.

Therefore,

$$\begin{gathered} ∆x=v_{1i}∆t+\frac{1}{2}\frac{F_{net}}{m}∆t^2..........\left(4\right) \\ \frac{d}{2}=\left(\begin{array}{c}{a}_{1}t\text{'}\end{array}\right)\left(\begin{array}{c}t\text{'}\text{'}\end{array}\right)........\left(5\right) \end{gathered}$$

However, the car's speed when it stops accelerating (at the end of this interval)

$$\begin{gathered} v_{1f}=\frac{∆x}{∆t} \\ =\frac{d}{t^{\text{'}}}..........\left(7\right) \end{gathered}$$

And

$$\begin{gathered} a_1=\frac{∆v}{∆t} \\ =\frac{V_{1f}}{t^{\text{'}}} \\ =\frac{d}{t^2}.......\left(8\right) \end{gathered}$$

Thus,

$\begin{array}{l}d=\frac{1}{2}\left(\frac{d}{t^2}\right)t^{\text{'}2}+\left(\frac{d}{t^{\text{'}2}}\right)t\text{'}{t}^{\text{'}\text{'}}......\left(9\right)\\ =\frac{d}{2}+\frac{d^{\text{'}\text{'}}}{t^{\text{'}}}......\left(10\right)\end{array}$

Thus,

$\frac{t^{\text{'}\text{'}}}{t^{\text{'}}}=\frac{1}{2}$

However, because $t=t^{\text{'}}+t^{\text{'}\text{'}}$the total time to travel the distance d is

$t=3t^{\text{'}\text{'}}$

$=\frac{3t^{\text{'}\text{'}}}{2}$(remember, car 1 and car 2 travel the same distance d in the same time t ).

Car 2, on the other hand, travels the entire distance under the effect of a net force.

As a result,

$$\begin{gathered} ∆x=v_{2i}t+\frac{1}{2}{a}_2∆t^{2}d........\left(11\right) \\ =\frac{1}{2}{a}_2{t}^2{a}_2........\left(12\right) \\ =\frac{2d}{t^2}.....\left(13\right) \\ butt=\frac{3t^{\text{'}}}{t^2},therefore, \\ {a}_2=\frac{8d}{9t^2}....\left(14\right) \end{gathered}$$

So the ratio of the initial acceleration of car 1 to that of car 2 is

$$\begin{gathered} \frac{a_1}{a_2}=\frac{d/t^{\text{'}2}}{8d/9t^{\text{'}2}} \\ =\frac{9}{8} \end{gathered}$$

The ratio of the final speed of car 1 to that of car 2.

(c)

For car 1 the final speed is

$$\begin{gathered} V_{1f}=V_{1j}+a_1{t}^{\text{'}} \\ =\frac{d}{t^{\text{'}}}........\left(1\right) \end{gathered}$$

Here the value of the acceleration have calculated in part (b), that is $a_1=\frac{d}{t^2}$.

Car has a final speed of (remember, from part (b)$a_2=\frac{8d}{9t^{\text{'}2}}$)

$$\begin{gathered} V_{2f}=V_{2j}+a_2∆t \\ =a_{2}t \\ =\frac{8d}{9t^{\text{'}2}}\times \frac{3}{2}{t}^{\text{'}} \\ =\frac{4d}{3t^{\text{'}}}..........\left(2\right) \end{gathered}$$

Therefore, the ratio of the final speed of car 1 to that of car 2 is

$$\begin{gathered} \frac{V_{1f}}{V_{2f}}=\frac{d/t^{\text{'}}}{4d/3t^{\text{'}}} \\ =\frac{3}{4} \end{gathered}$$