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NCERT Exemplar · Class 9 Science Motion

24 questions · 24 still being checked

Short Answer Questions 12–18 (part 2 of 3)

  1. Exercise 12

    The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero? Justify you answer.

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    NCERT’s answer
    No. Though the moving object comes back to its initial position the distance travelled is not zero.
    No. Displacement is the straight-line shift; distance is the actual path covered, and it can never be less than the displacement's size: \[d \ge |s| \] Example — a body goes from A to B and back to A along a straight road of length \(\displaystyle \ell\): \[s = (x_B-x_A)+(x_A-x_B) = 0 \] \[d = \ell+\ell = 2\ell \] Answer: No — distance is zero only when the object never moved; here \(\displaystyle d=2\ell\ne0\) although \(\displaystyle s=0\).

  2. Exercise 13

    How will the equations of motion for an object moving with a uniform velocity change?

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    NCERT’s answer
    Acceleration a = $\displaystyle 0$, v = u s = ut \(\displaystyle v^{2}\) - \(\displaystyle u^{2}\) = $\displaystyle 0$
    Set \(\displaystyle a=0\) in each equation of motion, since acceleration vanishes when velocity is uniform: \[v=u+at \;\xrightarrow{a=0}\; v=u \] \[s=ut+\tfrac12at^2 \;\xrightarrow{a=0}\; s=ut \] \[v^2=u^2+2as \;\xrightarrow{a=0}\; v^2=u^2 \] The first and third collapse to the trivial identity \(\displaystyle v=u\); only the second survives as a real relation. Answer: All three reduce to \(\displaystyle v=u\) and \(\displaystyle s=ut\) — distance equals uniform velocity times time; the third equation gives no new information.

  3. Exercise 14

    A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement-time graph is shown in Fig.8.4. Plot a velocity-time graph for the same. NCERT_Question_Class9_Science_Exemplar_Ch8_Q14

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Each segment of the displacement–time graph is a straight line, so the velocity is constant on each. \[v_1=\dfrac{\Delta s}{\Delta t}=\dfrac{100\ \text{m}-0}{50\ \text{s}-0}=2\ \text{m s}^{-1} \] \[v_2=\dfrac{\Delta s}{\Delta t}=\dfrac{0-100\ \text{m}}{100\ \text{s}-50\ \text{s}}=-2\ \text{m s}^{-1} \] Velocity is \(\displaystyle +2\ \text{m s}^{-1}\) walking to the letterbox and \(\displaystyle -2\ \text{m s}^{-1}\) walking back — same speed, opposite direction — reversing suddenly at \(\displaystyle t=50\ \text{s}\). NCERT_Solution_Class9_Science_Exemplar_Ch8_Q14 Answer: \(\displaystyle v=+2\ \text{m s}^{-1}\) for \(\displaystyle 0\)–\(\displaystyle 50\ \text{s}\); \(\displaystyle v=-2\ \text{m s}^{-1}\) for \(\displaystyle 50\)–\(\displaystyle 100\ \text{s}\).

  4. Exercise 15

    A car starts from rest and moves along the x-axis with constant acceleration 5\displaystyle 5 m s−2\displaystyle s^{-2} for 8\displaystyle 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12\displaystyle 12 seconds since it started from the rest?

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    NCERT’s answer
    The distance travelled in first $\displaystyle 8$ s, \(\displaystyle x_{1}\)= $\displaystyle 0$ + $\displaystyle 2$ ($\displaystyle 5$) ($\displaystyle 8$)$\displaystyle 2$ = $\displaystyle 160$ m. At this point the velocity v = u+ at = $\displaystyle 0$ + ($\displaystyle 5$×$\displaystyle 8$) =$\displaystyle 40$ m \(\displaystyle s^{-1}\) Therefore, the distance covered in last four seconds, \(\displaystyle x_{2}\) = ($\displaystyle 40$ × $\displaystyle 4$) m =$\displaystyle 160$ m Thus, the total distance x = \(\displaystyle x_{1}\)+\(\displaystyle x_{2}\) = ($\displaystyle 160$+ $\displaystyle 160$) m = $\displaystyle 320$ m
    Phase $\displaystyle 1$ — from rest, constant acceleration: \[v = u+at = 0+5\times8 = 40\ \text{m s}^{-1} \] \[s_1 = ut+\tfrac12at^2 = 0+\tfrac12(5)(8)^2 = 160\ \text{m} \] Phase $\displaystyle 2$ — the remaining \(\displaystyle 12-8=4\ \text{s}\) at this constant velocity: \[s_2 = vt_2 = 40\times4 = 160\ \text{m} \] Total distance since starting from rest: \[s = s_1+s_2 = 320\ \text{m} \] NCERT_Solution_Class9_Science_Exemplar_Ch8_Q15 Answer: \(\displaystyle 320\ \text{m}\).

  5. Exercise 16

    A motorcyclist drives from A to B with a uniform speed of 30\displaystyle 30 km h−1\displaystyle h^{-1} and returns back with a speed of 20\displaystyle 20 km h−1\displaystyle h^{-1}. Find its average speed.

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    NCERT’s answer
    Let AB = x, So \(\displaystyle t_{1}\)=, and = x x t Total time = \(\displaystyle t_{1}\) + \(\displaystyle t_{2}\) = $\displaystyle 5$ x h. -$\displaystyle 1$ Total distance Average speed for entire journey = = = $\displaystyle 24$ km h Total Time x x
    Speed is distance over time, so express each leg's time in terms of the common distance \(\displaystyle d\): \[t_1=\dfrac{d}{30},\qquad t_2=\dfrac{d}{20} \] \[t_{\text{total}} = \dfrac{d}{30}+\dfrac{d}{20} = \dfrac{2d+3d}{60} = \dfrac{5d}{60} = \dfrac{d}{12} \] \[v_{\text{avg}} = \dfrac{\text{total distance}}{\text{total time}} = \dfrac{2d}{d/12} = 24\ \text{km h}^{-1} \] Answer: \(\displaystyle 24\ \text{km h}^{-1}\).

  6. Exercise 17

    The velocity-time graph (Fig. 8.5\displaystyle 8.5) shows the motion of a cyclist. Find (i) its acceleration (ii) its velocity and (iii) the distance covered by the cyclist in 15\displaystyle 15 seconds. NCERT_Question_Class9_Science_Exemplar_Ch8_Q17

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    NCERT’s answer
    (i)
    Since velocity is not changing, acceleration is equal to zero. (ii) Reading the graph, velocity = $\displaystyle 20$ m \(\displaystyle s^{-1}\) (iii) Distance covered in $\displaystyle 15$ seconds, s = u × t = $\displaystyle 20$ × $\displaystyle 15$ = $\displaystyle 300$ m
    The graph is a horizontal line at \(\displaystyle 20\ \text{m s}^{-1}\): velocity never changes, so the slope, the acceleration, is zero. \[a=\dfrac{\Delta v}{\Delta t}=0 \] \[v=20\ \text{m s}^{-1}\ \text{(read directly off the graph)} \] \[s=vt=20\ \text{m s}^{-1}\times15\ \text{s}=300\ \text{m} \]Answer: (i) \(\displaystyle a=0\); (ii) \(\displaystyle v=20\ \text{m s}^{-1}\); (iii) \(\displaystyle s=300\ \text{m}\).

  7. Exercise 18

    Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height.

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    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    Taking upward as positive, gravity gives the same constant acceleration \(\displaystyle -g\) throughout the flight, so one equation covers the rise and the fall: \[v = u-gt \] At \(\displaystyle t=0\), \(\displaystyle v=u\) (launch); at \(\displaystyle t=u/g\), \(\displaystyle v=0\) (maximum height); at \(\displaystyle t=2u/g\), \(\displaystyle v=-u\) (back at the start, moving down). Plotting the speed \(\displaystyle |v|\) instead folds the falling half up into a V, the shape NCERT prints. NCERT_Solution_Class9_Science_Exemplar_Ch8_Q18 Answer: One line of slope \(\displaystyle -g\) through \(\displaystyle (0,u)\), \(\displaystyle (u/g,0)\), \(\displaystyle (2u/g,-u)\); its mirror above the axis after the peak is NCERT's speed graph.NCERT prints: a V-shaped graph labelled "Velocity" — a loose label: after the peak the signed velocity is negative, so the V is the speed \(\displaystyle |v|\), drawn here as the dashed curve.