Question

Two cars cover the same distance in a straight line. Car A covers the distance at a constant velocity. Car B starts from rest and maintains a constant acceleration. Both cars cover a distance of 491 m in 220 s. Assume that they are moving in the +x direction. Determine (a) the constant velocity of car A, (b) the final velocity of car B, and (c) the acceleration of car B.

Answer 1

Answer:

(a) $v_A=2.23m/s$

(b) $v_B=4.46m/s$

(c) $a=0.0202m/s^{2}$

Explanation:

The kinematic equation of car A is:

$(1): v_A=\frac{x}{t}$

On the other hand, the kinematic equations of car B are:

$(2):v_B=at\\\\(3):x=\frac{1}{2}at^{2}$

First, we can compute the constant velocity of car A from (1):

$v_A=\frac{x}{t} \\\\v_A=\frac{491m}{220s}=2.23m/s$

Next, we can obtain the acceleration of car B from (3):

$a=\frac{2x}{t^{2}} \\\\a=\frac{2(491m)}{(220s)^{2}}=0.0202m/s^{2}$

Finally, using the equation (2), we can get the final velocity of car B:

$v_B=at\\\\v_B=(0.0202m/s^{2})(220s)=4.46m/s$

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).