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Chapter 3: Problem 71

To mark a point for set of values \((1,5)\), look for 1 s on X-axis. Draw a line parallel to Y-axis and passing through this point. Look for \(5 \mathrm{~m}\) on the Y-axis and draw a line parallel to the X-axis passing through this point. The point of intersection gives the point that represents \((1,5) .\) In the same way, the points for other set of values can be plotted, as shown in the figure. (i) Join all the points. It is a straight line. The straight line is the distance-time graph for the motion of the motorbike. (ii) The distance-time graph of a body moving with a constant speed is a straight line. However, if the body does not move with constant speed, then its distance-time graph cannot be a straight line.

Short Answer

Expert verified
Answer: In a distance-time graph, a straight line indicates that the object is moving with a constant speed. The slope of the line represents the object's speed, and the points on the graph represent the distance traveled at different time intervals.

Step by step solution

01

Plot the point (1,5) on the graph

Locate the value 1 on the X-axis, which represents time. Draw a line parallel to the Y-axis, passing through this point. Next, locate the value 5 on the Y-axis, which represents distance. Draw a line parallel to the X-axis passing through this point. The point of intersection of these two lines represents the coordinate \((1,5)\), which we will plot on the graph.
02

Plot points for other set of values

Suppose we have other sets of values representing the time and distance traveled by the motorbike. Follow the same procedure as in Step 1 to plot these points on the graph. For example, if the set of values is \((2,10)\), look for 2 on the X-axis and 10 on the Y-axis, and plot the point accordingly. Repeat this process for all sets of values.
03

Join all the points and analyze the graph

After plotting all the points representing the motorbike's motion, connect these points with a straight line. Since the motorbike moves with constant speed, its distance-time graph will be a straight line. If the motorbike were not moving at a constant speed, the graph would not be a straight line. Conclusion: The distance-time graph of a motorbike (or any object) moving with a constant speed is a straight line. By plotting points and connecting them with a straight line, we can easily analyze the motion of the motorbike and determine its constant speed.

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Most popular questions from this chapter

\(\mathrm{A} \rightarrow \mathrm{b} \quad\) The piston of a motorcar engine moving at uniform speed is said to be in periodic motion. \(\mathrm{B} \rightarrow \mathrm{e}, \mathrm{b} \quad\) The objects executing vibratory motion undergo change in shape or size. The piston of a motor car engine executes vibratory motion. \(\mathrm{C} \rightarrow \mathrm{g} \quad\) Body at rest will have zero speed as well as zero velocity. \(\mathrm{D} \rightarrow\) a Maximum displacement of a body from its mean position is called amplitude. \(\mathrm{E} \rightarrow \mathrm{c} \quad\) A body moving with variable speed is said to be in non-uniform motion. \(\mathrm{F} \rightarrow \mathrm{d} \quad 1 \mathrm{~ms}^{-1}=\frac{1 \mathrm{~m}}{1 \mathrm{~s}}=\frac{\frac{1}{100} \mathrm{~km}}{\frac{1}{3600} \mathrm{~h}}=\frac{18}{5} \mathrm{~km} \mathrm{~h}^{-1}\) \(\mathrm{G} \rightarrow \mathrm{f} \quad\) Average velocity \(=\frac{\text { Total displacement }}{\text { Total time }}\)

From the given figure, by pythogorous theorem \(\mathrm{PR}^{2}=\mathrm{PQ}^{2}+\mathrm{R} \mathrm{Q}^{2}\) \(\mathrm{PR}^{2}=4^{2}+3^{2}\) \(\mathrm{PR}^{2}=16+9\) \(\mathrm{PR}=\sqrt{25}\) \(\mathrm{PR}=5 \mathrm{~m}\) Average speed \(=\frac{\text { total distance }}{\text { total time }}=\frac{4 \mathrm{~m}+3 \mathrm{~m}+5 \mathrm{~m}}{20 \mathrm{~s}+10 \mathrm{~s}+30 \mathrm{~s}}=\frac{12 \mathrm{~m}}{60 \mathrm{~s}}=0.2 \mathrm{~m} \mathrm{~s}^{-1}\)

Fill in the Blanks. length \(\mathrm{T} \alpha \sqrt{\ell}\) Time period of a simple pendulum depends on its length.

Fill in the Blanks. \(\frac{1}{1000}\) 1000 millisecond \(=1 \mathrm{~s}\) \(\Rightarrow 1\) millisecond \(=\frac{1}{1000}\) th part of a second.

Fill in the Blanks. vibratory The molecules in solid undergo vibratory motion.
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