The distance x (in μ m ) covered by a molecule starting from point A at time t=0 and stopping at another point B in given by the equation x=t^2 (2-(t)/(3)) The distance between A and B (in μ m ) is closed to

Correct answer: A. 10.7 Given that distance x=t^2(2-t / 3) On differentiating (d x)/(d t)=4 t-t^2 For x to be maximum, (d x)/(d t)=0 array rlr & 4 t-t^2=0 & t=4 array Maximum distance aligned x_ & =4^2(2-4 / 3) 16 & ×[(6-4)/(3)] = & 16 × (2)/(3) & =10.7 aligned

JIPMER 2015

Exam: JIPMER
Year: 2015
Subject: Physics
Chapter: Motion in One Dimension

JIPMER 2015 · Physics · Motion in One Dimension

The distance x (in μ m ) covered by a molecule starting from point...

The distance

x

(in

\mu \mathrm{m}

) covered by a molecule starting from point

A

at time

t=0 and stopping at another point

in given by the equation

x=t^2

\left(2-\frac{t}{3}\right)

The distance between

A

and

B

(in

\mu \mathrm{m}

) is closed to

Options

A10.7
B20.7
C40.7
D50.7

Correct answer

A. 10.7

Step-by-step solution

Given that distance

x=t^2(2-t / 3)

On differentiating

\frac{d x}{d t}=4 t-t^2

For

x

to be maximum,

\frac{d x}{d t}=0

\begin{array}{rlr} \Rightarrow & 4 t-t^2=0 \\ \Rightarrow & t=4 \end{array}

\therefore

Maximum distance

\begin{aligned} x_{\max } & =4^2(2-4 / 3) \\ 16 & \times\left[\frac{6-4}{3}\right] \\ = & 16 \times \frac{2}{3} \\ & =10.7\end{aligned}

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