Question

1. at t = 0, one toy car is set rolling on a straight track with initial position 15.0 cm, initial velocity - 3.50 cm/s, and constant acceleration 2.40 cm/s². at the same moment, another toy car is set rolling on an adjacent track with initial position 10.0 cm, initial velocity + 5.50 cm/s, and constant acceleration zero.

(a) at what time, if any, do 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?

Explanation:

Step1: Find the speed - time equation for each car

For the first car, the speed - time equation is $v_1 = v_{01}+a_1t$, where $v_{01}=- 3.50$ cm/s and $a_1 = 2.40$ cm/s². So $v_1=-3.50 + 2.40t$. For the second car, since $a_2 = 0$, the speed - time equation is $v_2=v_{02}$, and $v_{02}=5.50$ cm/s.

Step2: Solve for the time when speeds are equal

Set $v_1 = v_2$. So $-3.50+2.40t=5.50$.
Adding 3.50 to both sides gives $2.40t=5.50 + 3.50=9.00$.
Then $t=\frac{9.00}{2.40}=3.75$ s.

Step3: Find the speeds at that time

Since $v_2$ is constant, $v_1 = v_2=5.50$ cm/s at $t = 3.75$ s.

Step4: Find the position - time equations for each car

The position - time equation for the first car is $x_1=x_{01}+v_{01}t+\frac{1}{2}a_1t^2$, where $x_{01}=15.0$ cm, $v_{01}=-3.50$ cm/s and $a_1 = 2.40$ cm/s². So $x_1=15.0-3.50t + 1.20t^2$.
The position - time equation for the second car is $x_2=x_{02}+v_{02}t$, where $x_{02}=10.0$ cm and $v_{02}=5.50$ cm/s. So $x_2=10.0 + 5.50t$.

Step5: Solve for the time when the cars pass each other

Set $x_1=x_2$. So $15.0-3.50t + 1.20t^2=10.0 + 5.50t$.
Rearrange to get a quadratic equation $1.20t^2-9.00t + 5.00=0$.
Using the quadratic formula $t=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}$, where $a = 1.20$, $b=-9.00$ and $c = 5.00$.
$b^2-4ac=(-9.00)^2-4\times1.20\times5.00=81 - 24=57$.
$t=\frac{9.00\pm\sqrt{57}}{2\times1.20}=\frac{9.00\pm7.55}{2.40}$.
$t_1=\frac{9.00 + 7.55}{2.40}=\frac{16.55}{2.40}\approx6.89$ s and $t_2=\frac{9.00 - 7.55}{2.40}=\frac{1.45}{2.40}\approx0.604$ s.

Step6: Find the locations at those times

For $t = 0.604$ s, $x_2=10.0+5.50\times0.604=10.0 + 3.322 = 13.322$ cm.
For $t = 6.89$ s, $x_2=10.0+5.50\times6.89=10.0+37.895 = 47.895$ cm.

Answer:

(a) $t = 3.75$ s
(b) $v_1=v_2 = 5.50$ cm/s
(c) $t\approx0.604$ s and $t\approx6.89$ s
(d) At $t\approx0.604$ s, $x\approx13.3$ cm; at $t\approx6.89$ s, $x\approx47.9$ cm