Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
The displacement\[(x)\]of a particle depends on time\[(t)\,as\,x=\alpha {{t}^{2}}-\beta {{t}^{3}}\]. (1) The particle will come to rest after time\[\frac{2\alpha }{3\beta }\]. (2) The partice will return to its starting point after time\[\frac{\alpha }{\beta }\] (3) No net force will act on the particle at\[t=\frac{\alpha }{3\beta }\] (4) The initial velocity of the particle is not zeroA) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: A
Solution :
Displacement, \[x=a{{t}^{2}}-\beta {{t}^{3}}\] ... (i) \[v=\frac{dx}{dt}=2at-3\beta {{t}^{2}}\] ...(ii) Acceleration, \[a=\frac{dv}{dt}=2\alpha -6\beta t\] ... (iii) when particle will come to rest, then \[v=0\] From Eq. (ii), we get \[0=2\alpha t-3\beta {{t}^{2}}\] Or \[t=\frac{2\alpha }{3\beta }\] The particle when returns to its starting point, then. \[x=0\]. From Eq. (i), we get \[0=\alpha {{t}^{2}}-\beta {{t}^{3}}\] Or \[t=\frac{\alpha }{3\beta }\] Force on particle is zero, when\[a=0\] From Eq. (iii), we get \[0=2\alpha -6\beta t\] \[t=\frac{\alpha }{3\beta }\] At \[t=0\] \[v=2\alpha t-3\beta {{t}^{2}}\] \[v=0\] Hence, option is correct.
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