Question

After refueling, he starts from rest and leaves the pit area with an acceleration whose magnitude is 5.1 m/s2; after 3.6 s he enters the main speedway. At the same instant, another car on the speedway and traveling at a constant velocity of 71.3 m/s overtakes and passes the entering car. The entering car maintains its acceleration. How much time is required for the entering car to catch up with the other car?

Answer 1

Answer:

The entering car is going to catch up with the other car after 20.76 seconds.

Step-by-step explanation:

The car leaving the pit area is moving with a constant accelaration, then we can calculate the speed it has when entering the main speedway using the following equation:

$s_1=s_0+a*t_1$

Where $s_0$ is the initial speed (which is zero given that the car starts froms rest), $a$ is the car's accelaration and $t_1$ is the time passed until it reaches the main speedway.

$s_1=s_0+a*t_1=a*t_1$

$s_1=a*t_1=(5.1 \frac{m}{s^2})(3.6s)=18.36 \frac{m}{s}$

For an object travelling at a constant accelation, the displacement $x$ at a time $t$ can be calculate using the following equation:

$x=x_1+(s_1*t)+\frac{1}{2}at^2$ (equation 1)

Let's consider $x_1$ the point where the first car enters the main speedway. If $x_2$ is the point where this car catch up with other one after $t_2$ seconds, we may rewrite the equation 1 like this:

$x_2=x_1+(s_1*t_2)+\frac{1}{2}a(t_2)^2$

$x_2=(s_1*t_2)+\frac{1}{2}a(t_2)^2$ (equation 2)

In other hand, the second car is travelling at a constant speed $(v)$ when it meets the entering car. Then, its displacement $(d_2)$ after $t_2$ seconds can be calculated using the following formula:

$d_2=v*t_2$ (equation 3)

The entering car is going to catch up with the other one when $x_2=d_2$, so when can find how much time this will require equaling equations 2 and 3 and isolating $t_2$

$v*t_2=(s_1*t_2)+\frac{1}{2}a(t_2)^2$

$0=(s_1*t_2)+\frac{1}{2}a(t_2)^2-v*t_2$

$[\frac{1}{2}(a*t_2)-(v-s_1)]*t_2=0$

$\left \{ {{t_2=0} \atop {\frac{1}{2}(a*t_2)-(v-s_1)=0}} \right.$

$\frac{1}{2}(a*t_2)-(v-s_1)=0$

$t_2=\frac{2}{a}(v-s_1)=\frac{2}{5.1 \frac{m}{s^2}}(71.3\frac{m}{s}-18.36\frac{m}{s})=20.76 s$

So, the entering car is going to catch up with the other car after 20.76 seconds.