Answer:
x = ½ aw t²
x = d − ½ az t²
Explanation:
The position of each car at time t is:
x = x₀ + v₀ t + ½ at²
For car W:
x = 0 + (0) t + ½ aw t²
x = ½ aw t²
For car Z:
x = d + (0) t + ½ (-az) t²
x = d − ½ az t²
Solving the system of equations for x will give us the location the cars meet (relative to car W's starting point).
When the cars meet, the position on the horizontal surface is the difference in the distance traveled by each car [tex]X = d-\frac{1}{2} a_zt^2 \ \ or \ X = d - \frac{1}{2} a_wt^2[/tex]
The given parameters include;
acceleration of the Toy car W = [tex]a_w[/tex]
acceleration of Toy car Z, [tex]= a_z[/tex]
distance between both cars; = d
the initial velocity of both car, [tex]v_0 = 0[/tex]
The distance traveled by Toy car W;
[tex]d_w = v_0t + \frac{1}{2} a_wt^2\\\\d_w = \frac{1}{2} a_wt^2[/tex]
The distance traveled by Toy car Z;
[tex]d_z = v_0t + \frac{1}{2} a_zt^2\\\\d_z = \frac{1}{2} a_zt^2[/tex]
When the two cars meet, let the position on the horizontal = X;
[tex]X = d - d_z\\\\X = d - \frac{1}{2} a_zt^2 \\\\or\\\\X = d - d_w\\\\X = d - \frac{1}{2} a_wt^2[/tex]
Thus, when the cars meet, the position on the horizontal surface is the difference in the distance traveled by each car [tex]X = d-\frac{1}{2} a_zt^2 \ \ or \ X = d - \frac{1}{2} a_wt^2[/tex]
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