Question

Particle A starts from rest at t = 0 from x = 0 with constant acceleration to reach x = 1 m at t = 1 s. Particle B starts with uniform velocity at t = 0 from x = 1 to reach x = 2m at t = 1 s. What will be the distance covered by particle B in the frame of reference attached to particle A from t = 0 to t = 1 s. (upto two decimal places in meter)

• A. 0.50
• B. 0
• C. 0.25
• D. None of these

Answer 1

The correct option is A

0.50

$u=0,\frac{1}{2}{a}_A{t}^2=s\Rightarrow \frac{1}{2}{a}_A{1}^2\Rightarrow a_A=2m/s^2{v}_A=u_A+a_{A}t=2t{u}_B=?,a_B=0,ut+\frac{1}{2}a{t}^2=1\Rightarrow u_B=1m/s{v}_{BA}=1-2t,spee{d}_{BA}=|1-2t|,distance={\int }_0^t|1-2t|dt=\frac{1}{2}$