201. A \(v-t\) graph is a horizontal line below the time axis. The body is
ⓐ. moving in the positive direction with increasing speed
ⓑ. moving in the negative direction with constant speed
ⓒ. at rest because the graph is horizontal
ⓓ. moving with positive acceleration
Correct Answer: moving in the negative direction with constant speed
Explanation: A graph below the time axis means the velocity is negative. Negative velocity indicates motion in the negative direction. A horizontal \(v-t\) graph means the velocity is constant because it does not change with time. Constant velocity also means zero acceleration. The body is not at rest unless the horizontal line lies on the time axis where \(v=0\,\text{m s}^{-1}\).
202. A straight \(v-t\) graph crosses the time axis from positive velocity to negative velocity. The crossing point represents
ⓐ. an instant of maximum distance from the origin
ⓑ. an instant of zero acceleration necessarily
ⓒ. an instant of zero velocity and reversal
ⓓ. an instant of zero time interval
Correct Answer: an instant of zero velocity and reversal
Explanation: On a \(v-t\) graph, the time axis corresponds to \(v=0\,\text{m s}^{-1}\). If the graph crosses from above the axis to below it, velocity changes from positive to negative. At the crossing instant, the body has zero velocity. The sign change shows reversal of direction in one-dimensional motion. The crossing point does not by itself give position or distance from the origin.
203. A velocity-time graph passes through the points \(P(2\,\text{s},4\,\text{m s}^{-1})\) and \(Q(6\,\text{s},12\,\text{m s}^{-1})\). What is the acceleration represented by the straight line through these points?
ⓐ. \(+1\,\text{m s}^{-2}\)
ⓑ. \(+2\,\text{m s}^{-2}\)
ⓒ. \(+4\,\text{m s}^{-2}\)
ⓓ. \(+8\,\text{m s}^{-2}\)
Correct Answer: \(+2\,\text{m s}^{-2}\)
Explanation: \( \textbf{Graph points:} \) \(P(2\,\text{s},4\,\text{m s}^{-1})\) and \(Q(6\,\text{s},12\,\text{m s}^{-1})\).
On a \(v-t\) graph, acceleration is the slope:
\[
a=\frac{\Delta v}{\Delta t}
\]
Change in velocity:
\[
\Delta v=12\,\text{m s}^{-1}-4\,\text{m s}^{-1}=8\,\text{m s}^{-1}
\]
Change in time:
\[
\Delta t=6\,\text{s}-2\,\text{s}=4\,\text{s}
\]
Therefore,
\[
a=\frac{8\,\text{m s}^{-1}}{4\,\text{s}}
\]
\[
a=+2\,\text{m s}^{-2}
\]
The positive slope shows that velocity is increasing with time.
\( \textbf{Final answer:} \) The acceleration is \(+2\,\text{m s}^{-2}\).
204. A \(v-t\) graph lies below the time axis but slopes upward toward the axis. The body is
ⓐ. moving in the negative direction and slowing down
ⓑ. moving in the positive direction and speeding up
ⓒ. moving in the negative direction and speeding up
ⓓ. at rest because the graph is below the axis
Correct Answer: moving in the negative direction and slowing down
Explanation: A \(v-t\) graph below the time axis represents negative velocity, so the body is moving in the negative direction. An upward slope on a \(v-t\) graph represents positive acceleration. Here velocity and acceleration have opposite signs. Opposite signs mean the magnitude of velocity decreases, so the body slows down. If the graph reaches the time axis, the velocity becomes zero at that instant.
205. Consider the following statements about a \(v-t\) graph.
Statement I: The sign of velocity is decided by whether the graph is above or below the time axis.
Statement II: The slope of the graph gives acceleration.
Statement III: A horizontal \(v-t\) graph always represents rest.
ⓐ. I and II only
ⓑ. II and III only
ⓒ. I and III only
ⓓ. I, II, and III
Correct Answer: I and II only
Explanation: Statement I is true because points above the time axis represent positive velocity, while points below it represent negative velocity. Statement II is true because acceleration is the rate of change of velocity with time, which is the slope of the \(v-t\) graph. Statement III is false because a horizontal graph means constant velocity, not necessarily zero velocity. Rest is represented only by a horizontal line on the time axis. A horizontal line above or below the axis represents non-zero constant velocity.
206. Match the \(v-t\) graph feature with its meaning.
| Graph feature | Meaning |
|---|
| P. Graph above time axis | 1. Positive velocity |
| Q. Graph below time axis | 2. Negative velocity |
| R. Positive slope | 3. Positive acceleration |
| S. Zero slope | 4. Zero acceleration |
ⓐ. P-2, Q-1, R-4, S-3
ⓑ. P-1, Q-3, R-2, S-4
ⓒ. P-4, Q-2, R-3, S-1
ⓓ. P-1, Q-2, R-3, S-4
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: On a \(v-t\) graph, the vertical position of the graph tells the sign of velocity. A graph above the time axis means positive velocity, and a graph below it means negative velocity. The slope of the graph tells acceleration. A positive slope represents positive acceleration, while zero slope represents zero acceleration. This matching separates velocity sign from acceleration sign, which are read from different graph features.
207. A body has the following velocity-time data.
| Time | Velocity |
|---|
| \(0\,\text{s}\) | \(+3\,\text{m s}^{-1}\) |
| \(2\,\text{s}\) | \(+3\,\text{m s}^{-1}\) |
| \(4\,\text{s}\) | \(+3\,\text{m s}^{-1}\) |
The corresponding \(v-t\) graph is
ⓐ. a horizontal line above the time axis
ⓑ. a horizontal line below the time axis
ⓒ. a straight line sloping upward
ⓓ. a straight line crossing the time axis
Correct Answer: a horizontal line above the time axis
Explanation: The velocity is \(+3\,\text{m s}^{-1}\) at every listed instant. Since the velocity is positive, the graph lies above the time axis. Since the velocity does not change with time, the graph is horizontal. A horizontal \(v-t\) graph has zero slope, so the acceleration is zero. It does not cross the time axis because the velocity never becomes \(0\,\text{m s}^{-1}\).
208. A \(v-t\) graph is a straight line with equation \(v=10-2t\), where \(v\) is in \( \text{m s}^{-1} \) and \(t\) is in \( \text{s} \). At what time does the body reverse direction?
ⓐ. \(2\,\text{s}\)
ⓑ. \(5\,\text{s}\)
ⓒ. \(8\,\text{s}\)
ⓓ. \(10\,\text{s}\)
Correct Answer: \(5\,\text{s}\)
Explanation: \( \textbf{Velocity function:} \)
\[
v=10-2t
\]
A reversal in one-dimensional motion occurs when velocity changes sign.
The sign change happens at the instant when
\[
v=0
\]
Set the expression equal to zero:
\[
10-2t=0
\]
\[
2t=10
\]
\[
t=5\,\text{s}
\]
Before this instant, \(v\) is positive; after this instant, \(v\) becomes negative.
\( \textbf{Final answer:} \) The body reverses direction at \(t=5\,\text{s}\).
209. A straight \(v-t\) graph slopes downward and remains above the time axis for the interval shown. The acceleration is
ⓐ. positive, and the body speeds up
ⓑ. negative, and the body slows down
ⓒ. zero, and the body moves uniformly
ⓓ. negative, and the body must move in the negative direction
Correct Answer: negative, and the body slows down
Explanation: A downward slope on a \(v-t\) graph represents negative acceleration. Since the graph remains above the time axis, the velocity is positive throughout the interval. Positive velocity and negative acceleration have opposite signs. Opposite signs mean the speed decreases. The body is still moving in the positive direction while it slows down.
210. A velocity-time graph is a straight line passing through \(v=-6\,\text{m s}^{-1}\) at \(t=0\,\text{s}\) and \(v=0\,\text{m s}^{-1}\) at \(t=3\,\text{s}\). The acceleration is
ⓐ. \(+2\,\text{m s}^{-2}\)
ⓑ. \(-2\,\text{m s}^{-2}\)
ⓒ. \(+3\,\text{m s}^{-2}\)
ⓓ. \(-6\,\text{m s}^{-2}\)
Correct Answer: \(+2\,\text{m s}^{-2}\)
Explanation: \( \textbf{Initial velocity:} \) \(v_1=-6\,\text{m s}^{-1}\) at \(t_1=0\,\text{s}\).
\( \textbf{Final velocity:} \) \(v_2=0\,\text{m s}^{-1}\) at \(t_2=3\,\text{s}\).
Acceleration is the slope of the \(v-t\) graph:
\[
a=\frac{v_2-v_1}{t_2-t_1}
\]
Substitute the signed values:
\[
a=\frac{0\,\text{m s}^{-1}-(-6\,\text{m s}^{-1})}{3\,\text{s}-0\,\text{s}}
\]
\[
a=\frac{+6\,\text{m s}^{-1}}{3\,\text{s}}
\]
\[
a=+2\,\text{m s}^{-2}
\]
The positive acceleration makes the negative velocity less negative until it reaches zero.
\( \textbf{Final answer:} \) The acceleration is \(+2\,\text{m s}^{-2}\).
211. A \(v-t\) graph shows a line segment exactly along the time axis. During that segment, the body has
ⓐ. zero velocity
ⓑ. maximum positive velocity
ⓒ. negative acceleration necessarily
ⓓ. changing position uniformly
Correct Answer: zero velocity
Explanation: The time axis on a \(v-t\) graph corresponds to \(v=0\,\text{m s}^{-1}\). If the graph lies along this axis for a time interval, the velocity is zero throughout that interval. The body is at rest in the chosen frame during that segment. Since the velocity is constant at zero, the acceleration is also zero during that segment. The graph gives no uniform change of position because zero velocity means position is not changing.
212. The area under a velocity-time graph for straight-line motion gives
ⓐ. acceleration
ⓑ. displacement
ⓒ. final position coordinate directly
ⓓ. instantaneous speed only
Correct Answer: displacement
Explanation: A velocity-time graph has velocity \(v\) on the vertical axis and time \(t\) on the horizontal axis. The area under the graph has unit \((\text{m s}^{-1})(\text{s})=\text{m}\), which is the unit of displacement. If the graph is above the time axis, the area contributes positive displacement. If the graph is below the time axis, the area contributes negative displacement. This signed-area idea is different from slope, which gives acceleration on a \(v-t\) graph.
213. A body moves with constant velocity \(+6\,\text{m s}^{-1}\) for \(5\,\text{s}\). The displacement from the \(v-t\) graph is
ⓐ. \(+11\,\text{m}\)
ⓑ. \(+30\,\text{m}\)
ⓒ. \(+6\,\text{m}\)
ⓓ. \(+1.2\,\text{m}\)
Correct Answer: \(+30\,\text{m}\)
Explanation: \( \textbf{Graph shape:} \) A constant velocity gives a horizontal \(v-t\) graph.
\( \textbf{Velocity:} \) \(v=+6\,\text{m s}^{-1}\).
\( \textbf{Time interval:} \) \(t=5\,\text{s}\).
The area under the graph is a rectangle:
\[
\text{area}=v\times t
\]
\[
\text{area}=(+6\,\text{m s}^{-1})(5\,\text{s})
\]
\[
\text{area}=+30\,\text{m}
\]
Area under a \(v-t\) graph gives displacement, so
\[
s=+30\,\text{m}
\]
The positive sign appears because the graph is above the time axis.
\( \textbf{Final answer:} \) The displacement is \(+30\,\text{m}\).
214. A velocity-time graph is a horizontal line at \(v=-4\,\text{m s}^{-1}\) from \(t=0\,\text{s}\) to \(t=6\,\text{s}\). The displacement is
ⓐ. \(+24\,\text{m}\)
ⓑ. \(+10\,\text{m}\)
ⓒ. \(-10\,\text{m}\)
ⓓ. \(-24\,\text{m}\)
Correct Answer: \(-24\,\text{m}\)
Explanation: \( \textbf{Velocity:} \) \(v=-4\,\text{m s}^{-1}\).
\( \textbf{Time interval:} \) \(\Delta t=6\,\text{s}\).
For a horizontal \(v-t\) graph, the area is rectangular:
\[
\text{signed area}=v\Delta t
\]
\[
\text{signed area}=(-4\,\text{m s}^{-1})(6\,\text{s})
\]
\[
\text{signed area}=-24\,\text{m}
\]
The graph lies below the time axis, so the displacement is negative.
The magnitude \(24\,\text{m}\) is the path length if the velocity remains negative without reversal, but the signed displacement is \(-24\,\text{m}\).
\( \textbf{Final answer:} \) The displacement is \(-24\,\text{m}\).
215. A body starts from rest and its velocity increases uniformly to \(12\,\text{m s}^{-1}\) in \(4\,\text{s}\). The displacement from the \(v-t\) graph is
ⓐ. \(12\,\text{m}\)
ⓑ. \(24\,\text{m}\)
ⓒ. \(36\,\text{m}\)
ⓓ. \(48\,\text{m}\)
Correct Answer: \(24\,\text{m}\)
Explanation: \( \textbf{Graph shape:} \) Velocity increases uniformly from \(0\) to \(12\,\text{m s}^{-1}\), so the \(v-t\) graph is a triangle.
\( \textbf{Base of triangle:} \) \(4\,\text{s}\).
\( \textbf{Height of triangle:} \) \(12\,\text{m s}^{-1}\).
Area under the \(v-t\) graph gives displacement:
\[
s=\frac{1}{2}\times \text{base}\times \text{height}
\]
\[
s=\frac{1}{2}\times4\,\text{s}\times12\,\text{m s}^{-1}
\]
\[
s=24\,\text{m}
\]
The graph is above the time axis, so the displacement is positive.
Forgetting the factor \(\frac{1}{2}\) would give the rectangular area, not the triangular area.
\( \textbf{Final answer:} \) The displacement is \(24\,\text{m}\).
216. A \(v-t\) graph shows velocity increasing uniformly from \(4\,\text{m s}^{-1}\) to \(10\,\text{m s}^{-1}\) in \(3\,\text{s}\). The displacement is
ⓐ. \(12\,\text{m}\)
ⓑ. \(18\,\text{m}\)
ⓒ. \(21\,\text{m}\)
ⓓ. \(30\,\text{m}\)
Correct Answer: \(21\,\text{m}\)
Explanation: \( \textbf{Initial velocity:} \) \(u=4\,\text{m s}^{-1}\).
\( \textbf{Final velocity:} \) \(v=10\,\text{m s}^{-1}\).
\( \textbf{Time interval:} \) \(t=3\,\text{s}\).
The area under the graph is a trapezium:
\[
s=\frac{1}{2}(u+v)t
\]
\[
s=\frac{1}{2}(4+10)(3)
\]
\[
s=\frac{1}{2}(14)(3)
\]
\[
s=21\,\text{m}
\]
The same result can be seen as rectangle \(4\times3\) plus triangle \(\frac{1}{2}\times3\times6\).
Both parts are above the time axis, so the displacement is positive.
\( \textbf{Final answer:} \) The displacement is \(21\,\text{m}\).
217. A velocity-time graph lies above the time axis from \(0\,\text{s}\) to \(3\,\text{s}\) and below the time axis from \(3\,\text{s}\) to \(5\,\text{s}\). For finding displacement, the areas should be
ⓐ. added as positive areas only
ⓑ. subtracted using signed areas
ⓒ. ignored because the graph crosses the time axis
ⓓ. replaced by the slope of the graph
Correct Answer: subtracted using signed areas
Explanation: Displacement from a \(v-t\) graph is obtained from signed area. Area above the time axis is positive because velocity is positive there. Area below the time axis is negative because velocity is negative there. Therefore, the net displacement is found by adding the positive area and the negative area algebraically. If all areas are added as positive, the result gives distance travelled, not displacement.
218. Use the graph description below.
From \(t=0\,\text{s}\) to \(t=4\,\text{s}\), a \(v-t\) graph is a horizontal line at \(+5\,\text{m s}^{-1}\). From \(t=4\,\text{s}\) to \(t=7\,\text{s}\), it is a horizontal line at \(-2\,\text{m s}^{-1}\).
The displacement for the whole interval is
ⓐ. \(+14\,\text{m}\)
ⓑ. \(+20\,\text{m}\)
ⓒ. \(+26\,\text{m}\)
ⓓ. \(-26\,\text{m}\)
Correct Answer: \(+14\,\text{m}\)
Explanation: \( \textbf{First interval:} \) Velocity is \(+5\,\text{m s}^{-1}\) for \(4\,\text{s}\).
\[
s_1=(+5)(4)=+20\,\text{m}
\]
\( \textbf{Second interval:} \) Velocity is \(-2\,\text{m s}^{-1}\) for \(3\,\text{s}\).
\[
s_2=(-2)(3)=-6\,\text{m}
\]
Net displacement is the algebraic sum:
\[
s=s_1+s_2
\]
\[
s=+20\,\text{m}-6\,\text{m}
\]
\[
s=+14\,\text{m}
\]
The negative area reduces the displacement because the body moves in the negative direction during the second interval.
\( \textbf{Final answer:} \) The displacement is \(+14\,\text{m}\).
219. For the graph in which the positive area under a \(v-t\) curve is \(18\,\text{m}\) and the negative area has magnitude \(7\,\text{m}\), the displacement and distance are respectively
ⓐ. \(25\,\text{m}\) and \(11\,\text{m}\)
ⓑ. \(11\,\text{m}\) and \(25\,\text{m}\)
ⓒ. \(-11\,\text{m}\) and \(25\,\text{m}\)
ⓓ. \(18\,\text{m}\) and \(7\,\text{m}\)
Correct Answer: \(11\,\text{m}\) and \(25\,\text{m}\)
Explanation: \( \textbf{Positive area:} \) \(+18\,\text{m}\).
\( \textbf{Negative area:} \) \(-7\,\text{m}\).
Displacement is signed area:
\[
s=+18\,\text{m}-7\,\text{m}=+11\,\text{m}
\]
Distance is total absolute area:
\[
\text{distance}=18\,\text{m}+7\,\text{m}
\]
\[
\text{distance}=25\,\text{m}
\]
Displacement uses algebraic signs, while distance counts all covered path length positively.
\( \textbf{Final answer:} \) Displacement \(=+11\,\text{m}\), and distance \(=25\,\text{m}\).
220. A \(v-t\) graph is a straight line from \(v=+8\,\text{m s}^{-1}\) at \(t=0\,\text{s}\) to \(v=0\,\text{m s}^{-1}\) at \(t=4\,\text{s}\). The displacement is
ⓐ. \(8\,\text{m}\)
ⓑ. \(16\,\text{m}\)
ⓒ. \(24\,\text{m}\)
ⓓ. \(32\,\text{m}\)
Correct Answer: \(16\,\text{m}\)
Explanation: \( \textbf{Graph shape:} \) The velocity decreases linearly from \(+8\,\text{m s}^{-1}\) to \(0\,\text{m s}^{-1}\), so the region under the graph is a triangle.
\( \textbf{Base:} \) \(4\,\text{s}\).
\( \textbf{Height:} \) \(8\,\text{m s}^{-1}\).
Displacement is the area under the \(v-t\) graph:
\[
s=\frac{1}{2}\times4\times8
\]
\[
s=16\,\text{m}
\]
The graph is above the time axis throughout this interval, so the displacement is positive.
The body slows down, but it still keeps moving in the positive direction until \(v=0\,\text{m s}^{-1}\).
\( \textbf{Final answer:} \) The displacement is \(16\,\text{m}\).
