Answer:
a) 5.65 s
b) 5cm/s
c) They will pass each other at both 1.1168 s and 5.84s
d)15.084cm and 38.7 cm
Explanation:
For part A, you need to keep in mind that acceleration is the rate of change of velocity per unit of time. For a constant acceleration, this can be told in this way:
[tex]a = \frac{v - v_o}{t}[/tex]
Reordering this equation, we can get v in terms of the initial velocity, the acceleration, and the time elapsed:
[tex]v = at + v_o[/tex]
Now, we can get the expressions for velocity of each toy car, and equalize them:
[tex]v_1 =a_1t + v_o_1\\v_2 =a_2t + v_o_2\\v_1 = v2\\a_1t +v_o_1 =a_2t + v_o_2\\(a_1 - a_2)t = v_o_2 - v_o_1\\t = \frac{v_o_2 - v_o_1}{a_1 - a_2} = \frac{5cm/s - (-3cm/s)}{2.3 cm/s^2 - 0 cm/s^2}= 3.47 s[/tex]
As toy car has no acceleration and, therefore, constant speed, both car will have the same speed when toy car 1 reaches this velocity = 5cm/s
c) The position of car 1, as it follows a constant acceleration motion, is given by this equation:
[tex]x_1 = \frac{1}{2}a_1t^2 + v_o_1t + x_o_1[/tex]
The position for car 2, as it has constant velocity, is given by this equation:
[tex]x_2 = v_2t + x_o_2[/tex]
We equalize both equation to find the time where the cars pass each other:
[tex]x_1 = x_2\\\frac{1}{2}a_1t^2 + v_o_1t + x_o_1 = v_2t+x_o_2\\\frac{1}{2} a_1t^2 + (v_o_1 - v_2)t + x_o_1 - x_o_2 = 0\\\frac{1}{2}2.3m/s^2t^2 +(-3cm/s-5cm/s)t+ 17cm - 9.5cm = 0\\1.15t^2 -8t + 7.5 = 0 | a = 1.15, b = -8, c = 7.5\\t = \frac{-b +-\sqrt{b^2 - 4ac}}{2a} = 5.84s | 1.1168 s[/tex]
The car will pass each other at both 1.1168s and 5.84s.
For the positions, we solve any of the position equation with the solutions:
[tex]x = v_2*t + x_o_2 = 5cm/s *5.84s + 9.5cm = 38.7 cm\\x = 5cm/s * 1.1168s + 9.5cm = 15.084 cm[/tex]
