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 3 s distance travelled in 2 s As, u = 0, \[\,{{S}_{(3rd\,s)}}=\frac{1}{2}a{{.3}^{2}}-\frac{1}{2}a{{.2}^{2}}=\frac{1}{2}.a.5\] Given \[a=\frac{4}{3}m{{s}^{-2}}\] \[\therefore \,\,\,\,{{D}_{(3rd\,s)}}=\frac{1}{2}\times \frac{4}{3}\,\times 5=\frac{10}{3}\,m\]
You need to login to perform this action.
You will be redirected in
3 sec
