(B) From the figure,the position of $O$ is $0\; m$,$P$ is $360\; m$,and $Q$ is $240\; m$.Total displacement = Final position - Initial position = $240

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(B) From the figure,the position of $O$ is $0\; m$,$P$ is $360\; m$,and $Q$ is $240\; m$.Total displacement = Final position - Initial position = $240\; m - 0\; m = 240\; m$.Total time taken = $18\; s + 6.0\; s = 24\; s$.Average velocity = $\frac{\text{Total displacement}}{\text{Total time}} = \frac{240\; m}{24\; s} = 10\; m s^{-1}$.Total path length = Distance $OP$ + Distance $PQ = 360\; m + (360\; m - 240\; m) = 360\; m + 120\; m = 480\; m$.Average speed = $\frac{\text{Total path length}}{\text{Total time}} = \frac{480\; m}{24\; s} = 20\; m s^{-1}$.

Class 11 Physics Motion in Straight Line Mix Examples-Motion in Straight Line

Easy

$A$ car is moving along a straight line,say $OP$ in the given figure. It moves from $O$ to $P$ in $18\; s$ and returns from $P$ to $Q$ in $6.0\; s$. What are the average velocity and average speed of the car in going from $O$ to $P$ and back to $Q?$

A
$20\; m s^{-1}, 20\; m s^{-1}$
B
$10\; m s^{-1}, 20\; m s^{-1}$
C
$20\; m s^{-1}, 10\; m s^{-1}$
D
$30\; m s^{-1}, 10\; m s^{-1}$

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Motion in Straight Line

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Mix Examples-Motion in Straight Line

$A$ man walks on a straight road from his home to a market $2.5 \; km$ away with a speed of $5 \; km \; h^{-1}$. Finding the market closed,he instantly turns and walks back home with a speed of $7.5 \; km \; h^{-1}$. What is the average speed of the man over the interval of time $0$ to $40 \; min$?

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$A$ body of mass $m$ moving along a straight line covers half the distance with a speed of $2 \,ms^{-1}$. The remaining half of the distance is covered in two equal time intervals with a speed of $3 \,ms^{-1}$ and $5 \,ms^{-1}$ respectively. The average speed of the particle for the entire journey is

MediumKCET 2009

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What is reaction time? On what does the reaction time depend?

Medium

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$A$ car moves in the positive $Y$-direction with velocity $v$ proportional to the distance travelled $y$ as $v(y) \propto y^\beta$, where $\beta$ is a positive constant. The car covers a distance $L$ with average velocity $\langle v \rangle$ proportional to $L$ as $\langle v \rangle \propto L^{1/3}$. The constant $\beta$ is given as:

MediumTS EAMCET 2018

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If a car covers $2/5^{th}$ of the total distance with speed $v_1$ and $3/5^{th}$ of the total distance with speed $v_2$,then the average speed is:

Medium

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$A$ car is moving along a straight line,say $OP$ in the given figure. It moves from $O$ to $P$ in $18\; s$ and returns from $P$ to $Q$ in $6.0\; s$. What are the average velocity and average speed of the car in going from $O$ to $P$ and back to $Q?$

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$A$ man walks on a straight road from his home to a market $2.5 \; km$ away with a speed of $5 \; km \; h^{-1}$. Finding the market closed,he instantly turns and walks back home with a speed of $7.5 \; km \; h^{-1}$. What is the average speed of the man over the interval of time $0$ to $40 \; min$?

What is reaction time? On what does the reaction time depend?

If a car covers $2/5^{th}$ of the total distance with speed $v_1$ and $3/5^{th}$ of the total distance with speed $v_2$,then the average speed is:

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