Question

Two Cars are in a Race. Car a passes the finish line 2 seconds before of Car B. Car A passes the finish line with a velocity V more than Car B. Both start the race with acceleration of 3m/s^2(Car A) and 2m/s^2(Car B).Calculate the value of V.

Answer 1

Car A: u = 0, a = 3m/s², time = t, final velocity = v, distance =d
v = 0 + 3t = 3t
2ad =v² - u²
⇒ 2*3*d = (3t)²
⇒ 6d = 9t²
⇒ d = 1.5t²

Car B: u = 0, a = 2m/s², time = t+2, final velocity = v', distance = d
v' = 0 + 2(t+2) = 2t+4
2ad =v'² - u²
⇒ 2*2*d = (2t+4)²
⇒ 4d = 4t²+16t+16
⇒ d = t²+4t+4

So 1.5t² = t²+4t+4
⇒ 0.5t² - 4t - 4 = 0
⇒ t² - 8t - 8 = 0
⇒ t = 2(2+√6)s = 4+2√6 s

v = 3t = (12+6√6) m/s
v' = 2t+4 = 2(4+2√6)+4 = (12+4√6) m/s
V = v - v' = 2√6 m/s

Answer 2

s = u t + 1/2 a t²

first car:  s =  0 + 1/2 * 3 * t²
2nd car:  s = 0 + 1/2 * 2 * (t+2)²

From the two equations:    t² - 8 t - 8 = 0
                so     t = 4 + 2 √6   sec

Now :     v1  = 0 + 3 t
             v2  = 0 + 2 (t+2)

subtract:    V  =  t - 4 =  2 √6   m/sec