At the instant the traffic light turns green, an automobile starts with a constant acceleration a of 2.30 m/s^2. At the same instant a truck, traveling with a constant speed of 8.70 m/s, overtakes and passes the automobile?

(a) How far beyond the traffic signal will the automobile overtake the truck? (b) How fast will the automobile be traveling at that instant?

3 Answers
Oct 22, 2017

b) The car will be travelling at $17.40$ m/s, and
a) The car will travel $65.8$ m
by the time it catches the truck.

Explanation:

The two vehicles will be side-by-side as soon as they have the same average speed, 8.70 m/s. Then they will have travelled the same distance in the same amount of time.

In order to average 8.70 m/s, the car must reach a final speed twice that value:

${v}_{a v e} = \frac{1}{2} \left(v + {v}_{o}\right)$ where $v$ is the final speed

Which becomes ${v}_{a v e} = \frac{1}{2} v$ or $v = 2 {v}_{a v e}$

So, having proven that fact, we realize that the car must reach 17.40 m/s.

To determine how much time this takes, we use

$v = {v}_{o} + a \cdot t$

$17.40 = 0 + \left(2.30\right) \cdot t$

$t = \frac{17.40}{2.30} = 7.57$ s

In that time, the truck will have travelled

$d = v \cdot t = 8.70 \frac{m}{s} \cdot 7.57 s = 65.8 m$

Oct 22, 2017

(a) $d = 65.8 \text{ m}$

(b) ${v}_{\text{auto" = 17.4" m/s}}$

Explanation:

(a) How far beyond the traffic signal will the automobile overtake the truck?

The distance equation for the auto is:

$d = \frac{1}{2} a {t}^{2} \text{ [1]}$

The distane equation for the truck it:

$d = v t \text{ [2]}$

Solve equation [2] for t:

$t = \frac{d}{v} \text{ [2.1]}$

Substitute equation [2.1] into equation [1]:

$d = \frac{1}{2} a {\left(\frac{d}{v}\right)}^{2}$

Rewrite as a quadratic equal to zero:

$\frac{a}{2 {v}^{2}} {d}^{2} - d = 0$

Factor:

$d \left(\frac{a}{2 {v}^{2}} d - 1\right) = 0$

The two roots are:

$d = 0 \mathmr{and} d = 2 {v}^{2} / a$

Discard the d = 0 root and substitute $a = 2.30 {\text{ m/s}}^{2}$ and $v = 8.70 \text{ m/s}$ into the remaining root:

$d = 2 \left(8.70 {\text{ m/s")^2/(2.30" m/s}}^{2}\right)$

$d = 65.8 \text{ m}$

(b) How fast will the automobile be traveling at that instant?

Use the equation:

${v}_{\text{auto" = at" [3]}}$

Substitute equation [2.1] into equation [3]:

${v}_{\text{auto" = ad/v_"truck}}$

Substitute, $d = 65.8 \text{ m}$, a = 2.30" m/s"^2$, \mathmr{and}$v_"truck" = 8.70" m/s"

v_"auto" = ((2.30" m/s"^2)(65.8" m"))/(8.70" m/s")#

${v}_{\text{auto" = 17.4" m/s}}$

Oct 22, 2017

a.) $65.9 m$

b.) $17.4 \frac{m}{s}$

Explanation:

Using the kinematic equations, we can write two separate functions and set them equal to each other.

For the automobile with the constant acceleration:

$d = {v}_{0} t + \frac{1}{2} a {t}^{2}$

Since the automobile's initial velocity(${v}_{0}$) is zero ("an automobile starts..."), we are left with:

$d = \left(0\right) t + \frac{1}{2} \left(2.30 \frac{m}{s} ^ 2\right) {t}^{2} \implies d = \left(1.15 \frac{m}{s} ^ 2\right) {t}^{2}$

For the truck, we now we have a constant acceleration and can use a more basic equation for distance:

$d = v t \implies d = \left(8.70 \frac{m}{s}\right) t$

We can set these two equations equal to each other to find out at what time these two vehicles will be at the same (distance) from the traffic light.

$\left(8.70 \frac{m}{s}\right) t = \left(1.15 \frac{m}{s} ^ 2\right) {t}^{2}$

Using some basic arithmetic tools, we can solve for $t$:

$\frac{8.70 \frac{\cancel{m}}{\cancel{s}}}{1.15 \frac{\cancel{m}}{s} ^ \left(\cancel{2}\right)} t = \frac{\cancel{1.15 \frac{m}{s} ^ 2}}{\cancel{1.15 \frac{m}{s} ^ 2}} {t}^{2}$

$\left(7.565 s\right) \cancel{\frac{t}{t}} = \frac{{t}^{2}}{\cancel{t}}$

$7.565 s = t$

After about $7.57$ seconds, the automobile will have caught up to the truck.

We can now use this time to figure out how far and how fast each vehicle has traveled.
Since both will be at the same distance, we can use either equation; however, the truck's equation is easier to use:

a.)
$d = \left(8.70 \frac{m}{s}\right) t$
$d = \left(8.70 \frac{m}{\cancel{s}}\right) \left(7.57 \cancel{s}\right) = 65.859 m = 65.9 m$

For part b.) we will need to use a different kinematic equation to find the final velocity(${v}_{f}$) at that distance from the traffic light:

${v}_{f} = {v}_{0} + a t$

b.)
${v}_{f} = \left(0 \frac{m}{s}\right) + \left(2.30 \frac{m}{s} ^ 2\right) \left(7.57 s\right) = 17.411 \frac{m}{s} = 17.4 \frac{m}{s}$