toy car travels across a horizontal surface with an acceleration of after starting from rest. toy car travels across the same surface toward car with an acceleration of after starting from rest. car is separated from car by a distance . which of the following pairs of equations could be used to determine the location on the horizontal surface where the two cars will meet, and why?

To determine the location on the horizontal surface where two toy cars will meet, given that both start from rest and accelerate toward each other with known accelerations and are initially separated by a distance ddd, you can use kinematic equations for uniformly accelerated motion.

Setup

• Let the two cars be Car A and Car B.
• Car A starts at position x=0x=0x=0 and accelerates with acceleration aAa_AaA​.
• Car B starts at position x=dx=dx=d and accelerates toward Car A with acceleration aBa_BaB​.
• Both start from rest, so initial velocities vA0=0v_{A0}=0vA0​=0 and vB0=0v_{B0}=0vB0​=0.

Equations of Motion

For Car A moving from left to right:

xA=12aAt2x_A=\frac{1}{2}a_At^2xA​=21​aA​t2

For Car B moving from right to left toward Car A:

xB=d−12aBt2x_B=d-\frac{1}{2}a_Bt^2xB​=d−21​aB​t2

They meet when xA=xBx_A=x_BxA​=xB​, so:

12aAt2=d−12aBt2\frac{1}{2}a_At^2=d-\frac{1}{2}a_Bt^221​aA​t2=d−21​aB​t2

Rearranging:

12aAt2+12aBt2=d\frac{1}{2}a_At^2+\frac{1}{2}a_Bt^2=d21​aA​t2+21​aB​t2=d

12(aA+aB)t2=d\frac{1}{2}(a_A+a_B)t^2=d21​(aA​+aB​)t2=d

t2=2daA+aBt^2=\frac{2d}{a_A+a_B}t2=aA​+aB​2d​

t=2daA+aBt=\sqrt{\frac{2d}{a_A+a_B}}t=aA​+aB​2d​​

Once ttt is found, the meeting position xxx can be found by substituting back into either equation, for example:

x=12aAt2=aAaA+aBdx=\frac{1}{2}a_At^2=\frac{a_A}{a_A+a_B}dx=21​aA​t2=aA​+aB​aA​​d

Summary of Pair of Equations to Use

• Position of Car A: xA=12aAt2x_A=\frac{1}{2}a_At^2xA​=21​aA​t2
• Position of Car B: xB=d−12aBt2x_B=d-\frac{1}{2}a_Bt^2xB​=d−21​aB​t2
• Meeting condition: xA=xBx_A=x_BxA​=xB​

These equations are used because both cars start from rest and accelerate uniformly toward each other along the same horizontal line, so their positions as functions of time are quadratic in time. Setting their positions equal finds the time and location where they meet

. No information about collisions or elastic/inelastic interactions is needed here since the question only asks for the meeting point before any collision occurs

. The relative motion concept confirms that the distance between them decreases due to their accelerations toward each other

. Thus, the correct approach is to use the kinematic equations for each car and solve for ttt and xxx where their positions coincide.