Answer :
Answer:
(a) [tex]v_A=2.23m/s[/tex]
(b) [tex]v_B=4.46m/s[/tex]
(c) [tex]a=0.0202m/s^{2}[/tex]
Explanation:
The kinematic equation of car A is:
[tex](1): v_A=\frac{x}{t}[/tex]
On the other hand, the kinematic equations of car B are:
[tex](2):v_B=at\\\\(3):x=\frac{1}{2}at^{2}[/tex]
First, we can compute the constant velocity of car A from (1):
[tex]v_A=\frac{x}{t} \\\\v_A=\frac{491m}{220s}=2.23m/s[/tex]
Next, we can obtain the acceleration of car B from (3):
[tex]a=\frac{2x}{t^{2}} \\\\a=\frac{2(491m)}{(220s)^{2}}=0.0202m/s^{2}[/tex]
Finally, using the equation (2), we can get the final velocity of car B:
[tex]v_B=at\\\\v_B=(0.0202m/s^{2})(220s)=4.46m/s[/tex]
In words, we got that the constant velocity of car A is 2.23m/s (a), the final velocity of car B is 4.46m/s (b), and the acceleration of car B is 0.0202m/s² (c).