A) 6 m
B) 4 m
C) \[\frac{10}{3}m\]
D) \[\frac{19}{3}m\]
Correct Answer: C
Solution :
| Distance travelled by the particle in nth second is\[{{S}_{nth}}=u+\frac{1}{2}a(2n-1)\] where u is initial speed and a is acceleration of the particle. |
| Here, \[n=3,u=0,a=\frac{4}{3}m/{{s}^{2}}\] |
| \[\therefore \] \[{{S}_{3rd}}=0+\frac{1}{2}\times \frac{4}{3}\times (2\times 3-1)\] |
| \[=\frac{4}{6}\times 5\] |
| \[=\frac{10}{3}m\] |
| Alternative: Distance travelled in the 3rd Second = distance travelled in 3s distance travelled in 2s |
| As, \[u=0,\] |
| \[\,{{S}_{(3rd\,s)}}=\frac{1}{2}a{{.3}^{2}}-\frac{1}{2}a{{.2}^{2}}=\frac{1}{2}.a.5\] |
| Given \[a=\frac{4}{3}m{{s}^{-2}}\] |
| \[\therefore \] \[{{D}_{(3rd\,s)}}=\frac{1}{2}\times \frac{4}{3}\,\times 5=\frac{10}{3}\,m\] |
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