A) 60 m
B) 85 m
C) 90 m
D) 100 m
Correct Answer: C
Solution :
Now, using the relation from equation of motion, \[{{s}_{t}}=u+\frac{1}{2}a(2t-1)\] \[{{s}_{t}}=u+a(t-\frac{1}{2})\] (Given, \[{{s}_{t}}=30m,\,t=3s,\,u=5m/s\]) Hence, \[30=5+a\left( 3-\frac{1}{2} \right)\] So, \[a=(30-5)\times \frac{2}{5}=10m/{{s}^{2}}\] Therefore, \[{{s}_{4}}=5+10\left( 4-\frac{1}{2} \right)\] \[=5+10\times \frac{7}{2}=40m\] and \[{{s}_{5}}=5+10\left( 5-\frac{1}{2} \right)\] \[=5+\frac{10\times 9}{2}=50m\] Hence, the distance covered after two second is \[s={{s}_{4}}+{{s}_{5}}=40+50=90m\]You need to login to perform this action.
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