A) x = 3y
B) y = 3x
C) x = y
D) y - 2x
Correct Answer: B
Solution :
Key Idea: Since car accelerates from rest initial velocity is zero. From equation of motion, we have \[s=ut+\frac{1}{2}a{{t}^{2}}\]where u is initial velocity, t is time and a is acceleration. Since car accelerates from rest u = 0, t = 10s \[\therefore \] \[s=0+\frac{1}{2}\times a\times {{(10)}^{2}}=50a\] ?(i) Also, v = u + at where, v is final velocity. \[\therefore \] Velocity after 10 s is \[v=0+a\times 10\] \[v=10a=10\times \frac{s}{50}\] ?(ii) In the next 10s car moves with constant acceleration and with initial velocity v. \[\therefore \] \[s'=vt+\frac{1}{2}a{{t}^{2}}\] \[=\frac{s}{50}\times 10\times 10+\frac{1}{2}\times \frac{s}{50}\times 100=3s\] Given, \[s=x\]and \[s'=y\] \[\therefore \] \[y=3x\]
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