Question

41. Car A, moving in a straight line at a constant speed of 20. meters per second, is initially 200 meters behind car B , moving in the same straight line at a constant speed of 15 meters per second. how far must car a travel from this initial position before it catches up with car B?
(1) 200m
(2) 400m
(3) 800m
(4)1000m

Answer 1

800meters. Car A at 20 meters per second will travel 800 meters in 40 seconds. Car B in the same 40 seconds will only travel 600 meters at 15 meters per second. therefore car A catches up to car B after traveling 800 meters.

Answer 2

Answer:

(3) 800m

Explanation:

To solve this problem you just have to find first the time that it takes the car to catch up with car B, and you just have to create a simple ecuation, since distance= time*velocity

You know that the distance to catch up is 200 m, and both cars have different speed, so you just substract from the fastest car, the slowest car speed.

Speed= 20 $\frac{m}{s}$ -15$\frac{m}{s}$= 5 $\frac{m}{s}$

So now you just divide:

T=$\frac{d}{s}$

T= $\frac{200m}{[tex]5\frac{m}{s}$}[/tex]

T= 40 s

Now you just multiply the time for the speed of the first car:

D=SxT

D= $(20\frac{m}{s})(5s)$

D=800m