When Do Two Toy Cars on Parallel Tracks Match Speeds and Positions?

[BeckyStar678]

Homework Statement

at t=0, one toy car is set rolling on a straight track with initial position 15 cm, initial velocity -3.50 cm/s, and constant acceleration 2.40 cm/s2. at the same moment, another car is set rolling n an adjacent track with inital position 10 cm, an initial velocity of +5.50 cm/s, and constant acceleration zero. a.) at what time,if anydo the two cars have equal speeds b.) what are their speeds at that time? c.) at what time(s) if any do the cars pass each other? d.) what are their locations at that time? e.) explain the difference bt question a and c as clearly as possible

Homework Equations

x_f=x_i+ v_xit+1/2a_xt²

The Attempt at a Solution

is the above equation the right one to use?

 

[Doc Al]

That equation describes distance as a function of time. You'll need it, but you'll need others also. Review the https://www.physicsforums.com/showpost.php?p=905663&postcount=2".

 

[baba89]

Yes, the above equation is correct for solving this problem. It is the formula for calculating position (xf) at a given time (t) with initial position (xi), initial velocity (vxi), and constant acceleration (ax).

a) To determine when the two cars have equal speeds, we can set their velocity equations equal to each other and solve for t:

-3.50 + 2.40t = 5.50 + 0t (since the second car has zero acceleration)
2.40t = 9
t = 3.75 seconds

b) At this time, both cars have the same speed of 5.50 cm/s.

c) To determine when the cars pass each other, we can set their position equations equal to each other and solve for t:

15 + (-3.50)t + 1/2(2.40)t^2 = 10 + (5.50)t + 0t^2
2.40t^2 + 9t - 5 = 0
Using the quadratic formula, we get t = 0.44 seconds or t = -3.94 seconds. Since time cannot be negative, the cars pass each other at t = 0.44 seconds.

d) To find their locations at this time, we can plug t = 0.44 seconds into either of their position equations:

Car 1: x = 15 + (-3.50)(0.44) + 1/2(2.40)(0.44)^2 = 14.3 cm
Car 2: x = 10 + (5.50)(0.44) + 0(0.44)^2 = 12.4 cm

e) The difference between questions a and c is that in question a, we are looking for when the cars have equal speeds, meaning their velocities are equal at that time. In question c, we are looking for when the cars pass each other, meaning their positions are equal at that time. These two times may not necessarily be the same, as we can see in this problem where the cars have equal speeds at 3.75 seconds but pass each other at 0.44 seconds.