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

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

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 (m)/(s^2))(3.6s)=18.36 (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)+(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)+(1)/(2)a(t_2)^2`

`x_2=(s_1*t_2)+(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)+(1)/(2)a(t_2)^2`

`0=(s_1*t_2)+(1)/(2)a(t_2)^2-v*t_2`

`[(1)/(2)(a*t_2)-(v-s_1)]*t_2=0`

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

`(1)/(2)(a*t_2)-(v-s_1)=0`

`t_2=(2)/(a)(v-s_1)=(2)/(5.1 (m)/(s^2))(71.3(m)/(s)-18.36(m)/(s))=20.76 s`

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