How do I solve this 1-D Kinematic Question involving two cars?

Question

Two cars are heading towards one another. Car A is moving with an acceleration of a_A = 11 m/s². Car B is moving with an acceleration of a_B = -12 m/s². The cars are initially at rest and separated by a distance d = 2300 m on the x-axis.

a) At what time in seconds do the cars meet?

b) What is the displacement in meters of Car A and then Car B.

c) Now, assume each car starts with an initial velocity. Car A begins with an initial velocity of 23 m/s and Car B begins with an initial velocity of -11 m/s. What time in seconds do they meet?

Answer 1

I will set this up for you, but you do the math.

(a)

You need to find the time when they are both at the same place. Assuming car A is initially at x = 0 and car B is initially at x = 2300,

x_A(t) = a_At²/2 = 11t²/2 = 5.5t²

x_B(t) = 2300 + a_Bt²/2 = 2300 - 12t²/2 = 2300 - 6t²

Set x_A(t) = x_B(t) and solve for t.

5.5t² = 2300 - 6t²

You should be able to solve for t from here. The result will be in seconds.

(b)

Plug the result for t from part (a) into x_A(t) and x_B(t). The displacements will be

d_A = final position minus initial position = x_A(t) - 0 = x_A(t)

d_B = final position minus initial position = x_B(t) - 2300

(c)

Solve the same way as part (a), but use these equations for x_A and x_B (which account for the initial speeds):

x_A(t) = v_At + a_At²/2 = 23t + 5.5t²

x_B(t) = 2300 + v_Bt + a_Bt²/2 = 2300 - 11t - 6t²

Now set x_A(t) = x_B(t) and solve for t.

23t + 5.5t² = 2300 - 11t - 6t²

You should be able to solve for t and finish from here.