How Far and How Long Until the Sports Car Catches Up with the Truck?

[Dough]

A truck is traveling at a constant velocity of 22m/s north. The Driver sees a traffic light turn from red to green soon enough, so he does not have to alter his speed. Meanwhile, a woman in a sports car is stopped at a red light. At the moment the light turns green and the truck passes her, she begins to accelerate at 4.8[m/s]/s (metres per second squared). How far have both vehicles traveled when the sports car catches up with the truck? How long did it take for the sports car to catch up with the truck?

Thanks
-Dough

 

[Pengwuino]

A truck is traveling at a constant velocity of 22m/s north. The Driver sees a traffic light turn from red to green soon enough, so he does not have to alter his speed. Meanwhile, a woman in a sports car is stopped at a red light. At the moment the light turns green and the truck passes her, she begins to accelerate at 4.8[m/s]/s (metres per second squared). How far have both vehicles traveled when the sports car catches up with the truck? How long did it take for the sports car to catch up with the truck?

Thanks
-Dough

Ok so you want to find an equation for both of their positions at any given time. The trucks equation will be x(t)= 22t
the sports cars equation will be x(t)=.5at^2 which = 2.4t^2
Set the equations equal to each other so that its 22t=2.4t^2 and you find the 0's
One will be 0 but the other one will be the answer which is t=9.17 seconds which is more accurately defined as 55/6.
It will take 55/6 seconds to catch up. Plug 55/6 into either equation to find out how far both vehicles have traveled which is 201.67 meters or more exactly, 605/3

 

[Dough]

now i feel really stupid cuase i typed out the wrong question

ok that was easier than how the book explained it

the question that iw as supposed to type up is:

In a long distance race, Michael is running at 3.8 m/s and 75m behind Robert, who is running at a constant velocity of 4.2 m/s. If Michael accelerates at 0.15 m/s^2, how logn will it take him to catch Robert?

Damn i feel stupid for typing the wrong quesiton... what a first post...

 

[Pengwuino]

Lol ok well then, let's see now. Basically you set up 2 equations again
Michael: x(t)= -75 + 3.8t + 0.075t^2
Robert: x(t)= 4.2t

Find out the zeros obviously greater then 0.

The answer comes out to be 34.4s. This is verifyable by plugging it into both equations and the distance they meet at if the origin is 0 at Robert's starting point is about 144.5 meters

 

[Dough]

Lol ok well then, let's see now. Basically you set up 2 equations again
Michael: x(t)= -75 + 3.8t + 0.075t^2
Robert: x(t)= 4.2t

Find out the zeros obviously greater then 0.

The answer comes out to be 34.4s. This is verifyable by plugging it into both equations and the distance they meet at if the origin is 0 at Robert's starting point is about 144.5 meters

im going to give that a shot in a few minutes

Why isn't it

Michael: x(t)= 3.8t + 0.075t^2
Robert: x(t)= 75+ 4.2t

 

[Pengwuino]

It doesn't matter. When you put the 2 equations equal to each other, that 75 can be positive for robert or, as i put it, negative for michael.

 

[Dough]

Great! I got it, Thanks a lot Pengwuino(cool name too)

Ok before i was doing it like this

4.2m/s × t = -75m + 3.8m/s × t + 0.075t^2
because of this i kept trying to do something stupid... like

4.2m/s = -75m + 3.8m/s × t + (0.075t^2) / t
and then cancel out the power to the 2 but then units wouldn't really allow for me to do much...

Would all questions with an offset liekt his oen be quadratic? I did a lot of other questions and none required use of the quadratic formula...

 

[Pengwuino]

Only if there's a t^2 in it.