A) \[x=3y\]
B) \[y=3x\]
C) \[x=y\]
D) \[y=2x\]
Correct Answer: B
Solution :
From equation of motion, we have \[s=ut+\frac{1}{2}a{{t}^{2}}\] where \[u\] is initial velocity, \[t\] is time, \[a\] 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 \[10\,\,s\] car moves with constant acceleration and with initial velocity \[\therefore \] \[s=vt+\frac{1}{2}a{{t}^{2}}\] \[=\frac{s}{50}\times 10\times 10+\frac{1}{2}\times \frac{s}{50}\times 100=3s\] Given, \[s=x\]and\[s=y\] \[\therefore \] \[y=3x\]
You need to login to perform this action.
You will be redirected in
3 sec
