A) 5 m
B) 1 m
C) 2 m
D) 4 m
Correct Answer: D
Solution :
[d] Idea In expressions \[x=4t-2{{t}^{2}}\] the \['x'\] represents displacement not distance. |
\[x=4t-2{{t}^{2}},v=4(1-t)\] and at \[t=1,v=0\] |
The velocity of the particle become zero at \[t=1\text{ }s\] and then it moves towards the negative direction. |
\[\Rightarrow \] \[{{x}_{1}}\] for 1st second \[=4-2\times 1=+2\,m\] |
\[\Rightarrow \] \[{{x}_{2}}\] for 2nd second |
\[=4\times 2-2\times 4=8-8=0\] |
So, first particle has covers 2m and then it comes back 2 m. |
So, the total distance covered \[=2+2=4\text{ }m\]. |
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