Car A and B start moving simultaneously in the same direction along the line joining them. Car A with a constant acceleration a=4m/s2, while car B moves with a constant velocity v=1m/s. At time t=0, car A is 10m behind car B. The time when car A overtakes car B is: - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

Given,

u_A = 0, u_B = 1m/s, a_A = 4 m/s², a_B = 0

Assuming car B to be at rest, we have,
u_AB= u_A- u_B = 0-1 = -1 m/s
_aAB= a_A- a_B = 4-0 = 4 m/s²

$s = ut + \frac{1 }{2} {at}^{2}$
$10 = - t + \frac{1 }{2} (4) t^{2}$

$\Rightarrow 2 t^{2} - t - 10 = 0$

Solving, we get,

t= 2.5 or t= -2

Since time never is negative, t = 2.5 s is the correct response.

Car A and B start moving simultaneously in the same direction along the line joining them. Car A with a constant acceleration a=4m/s², while car B moves with a constant velocity v=1m/s. At time t=0, car A is 10m behind car B. The time when car A overtakes car B is: