Acceleration, distance, time. help.

[Eagleized]

Homework Statement

vehicle 1 is traveling at a constant speed of 90mph (132 f/s) 5 seconds before vehicle 2 begins to accelerate from 0-90 mph taking 17.09 seconds to do so. how far a distance will vehicle 2 travel while accelerating to 90mph. and how long will it take for vehicle 2 to catch up to vehicle 1 while accelerating at approximately 5.2 m/s/s or 15.6 f/s

Homework Equations

The relationship between the 2 vehicles is dependant.

what is the total distance traveled by vehicle 2 before it catches up to vehicle 1?

The Attempt at a Solution

ok, it's too hard to explain and argue at the same time, so i'll leave it at this:
find a function of position for car #1 and car #2, call them c_1(t) and c_2(t), set them equal and solve for t. what does this mean exactly?
This should help you a bit, at constant acceleration, call it a, the position of car 2 at time t is at^2/2
the first car, meanwhile is moving at a constant velocity v, so its position at time t is vt
does car 2 start from rest?
so solve vt=at^2/2 -> at^2-vt=0 -> use quadratic formula
this will give you the time t, when car 2 catches up to car 1, ?? to answer part 2 of the Q

 

[jgens]

For your first question you could use the equation Vf^2 = Vi^2 + 2(a)(d) where Vf is the final velocity, Vi is the initial velocity, a is the acceleration, and d is the displacement.

For number 2, since the first car is traveling at constant speed we can label the distance it travels as d = V(t + 5) while we may use the equation df = (a)(t^2)/2 + (Vi)t + di. Equate the two and solve for the time.

 

[Eagleized]

did you get a distance of 4044.8 ft for the first part of the question?

or 558.46 ft?

 

[Eagleized]

anyone get an answer for this?