question_answer

The position of a particle vary \[x=4t-2{{t}^{2}}\]. The distance covered by particle in 4 s is

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 \[=2+2=4\text{ }m\].

TEST Edge There could be different relations possible i.e., \[v=2x+5\] and one can ask the relation between x and t from the given equation. Here, put \[v=\frac{dx}{dt}\] and integrate the expression to get the relation between x and t.