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 460 m in 210 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.

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aristeus aristeus · Answer 1 · 2019-10-05T10:46:48+02:00

Answer:

(a) Constant velocity of car A is 2.19 m/sec

(b) Final velocity of car B is 4.62 m/sec

(c) Acceleration of car B is 0.022 [tex]m/sec^2[/tex]

Explanation:

We have given that both car travels a distance of 460 m in 210 sec

So distance s = 460 m and time t = 210 sec

(a) AS car A is traveling with constant velocity

So velocity of car A [tex]=\frac{distance}{time}=\frac{460}{210}=2.19m/sec[/tex]

(c) As car starts from rest its initial velocity u= 0 m/sec

According to second equation of motion [tex]s=ut+\frac{1}{2}at^2[/tex]

[tex]460=0\times 210+\frac{1}{2}\times a\times 210^2[/tex]

[tex]a=0.022m/sec^2[/tex]

(b) Now from first equation of motion

v=u+at

v=0+0.022×210 = 4.62 m/sec