201. For a non-zero displacement \(\Delta\vec{r}\) during a positive time interval \(\Delta t\), the direction of \(\vec{v}_{\text{avg}}\) is:
ⓐ. opposite to \(\Delta\vec{r}\)
ⓑ. perpendicular to \(\Delta\vec{r}\)
ⓒ. the same as \(\Delta\vec{r}\)
ⓓ. always along the positive \(x\)-axis
Correct Answer: the same as \(\Delta\vec{r}\)
Explanation: Average velocity is defined by \(\vec{v}_{\text{avg}}=\frac{\Delta\vec{r}}{\Delta t}\). The time interval \(\Delta t\) is a positive scalar for ordinary motion records. Dividing a vector by a positive scalar changes its magnitude but not its direction. Therefore, average velocity points in the same direction as displacement. The path taken between the two points does not decide this direction; the net displacement does.
202. A particle covers a total path length of \(100\,\text{m}\) in \(20\,\text{s}\), but its displacement is \(60\,\text{m}\) east. Its average speed and average velocity are:
ⓐ. \(5\,\text{m s}^{-1}\) and \(3\,\text{m s}^{-1}\) east
ⓑ. \(3\,\text{m s}^{-1}\) and \(5\,\text{m s}^{-1}\) east
ⓒ. both \(5\,\text{m s}^{-1}\) east
ⓓ. both \(3\,\text{m s}^{-1}\) without direction
Correct Answer: \(5\,\text{m s}^{-1}\) and \(3\,\text{m s}^{-1}\) east
Explanation: \( \textbf{Total distance:} \) \(100\,\text{m}\).
\( \textbf{Displacement:} \) \(60\,\text{m}\) east.
\( \textbf{Time interval:} \) \(20\,\text{s}\).
\( \textbf{Average speed:} \)
\[
\text{average speed}=\frac{\text{total distance}}{\Delta t}
\]
\[
\text{average speed}=\frac{100\,\text{m}}{20\,\text{s}}=5\,\text{m s}^{-1}
\]
\( \textbf{Average velocity magnitude:} \)
\[
|\vec{v}_{\text{avg}}|=\frac{|\Delta\vec{r}|}{\Delta t}
\]
\[
|\vec{v}_{\text{avg}}|=\frac{60\,\text{m}}{20\,\text{s}}=3\,\text{m s}^{-1}
\]
\( \textbf{Direction:} \) Average velocity has the direction of displacement, which is east.
\( \textbf{Final answer:} \) The average speed is \(5\,\text{m s}^{-1}\), and the average velocity is \(3\,\text{m s}^{-1}\) east.
203. Instantaneous velocity in a plane is best described as:
ⓐ. average velocity over an infinitesimal interval
ⓑ. the total distance divided by total time for the whole path
ⓒ. the product of displacement and time interval
ⓓ. the velocity found only at the final point of a journey
Correct Answer: average velocity over an infinitesimal interval
Explanation: Average velocity is calculated over a finite time interval using \(\vec{v}_{\text{avg}}=\frac{\Delta\vec{r}}{\Delta t}\). Instantaneous velocity is obtained when this interval is made very small around a particular instant. In calculus notation, it is written as \(\vec{v}=\frac{d\vec{r}}{dt}\). Since \(\vec{r}\) is a vector, instantaneous velocity is also a vector. It gives both the speed and direction of motion at that instant, not over the whole journey.
204. At a point on a smooth curved path, the instantaneous velocity of a moving body is directed:
ⓐ. toward the origin of the coordinate system
ⓑ. along the chord joining the starting and ending points
ⓒ. always along the positive \(x\)-axis
ⓓ. along the tangent to the path at that point
Correct Answer: along the tangent to the path at that point
Explanation: Instantaneous velocity tells the direction in which the body is moving at that exact point. On a curved path, the small displacement over a very short time becomes nearly tangent to the curve. In the limiting case, the direction of velocity is the tangent direction. The chord between two distant points gives an average displacement direction, not the instantaneous direction. This is why velocity direction can change continuously even when the path is smooth.
205. The position vector of a body is \(\vec{r}(t)=2t^2\,\hat{i}+(3t+1)\hat{j}\), where \(\vec{r}\) is in \(\text{m}\) and \(t\) is in \(\text{s}\). Its instantaneous velocity at \(t=2\,\text{s}\) is:
ⓐ. \(4\,\text{m s}^{-1}\hat{i}+3\,\text{m s}^{-1}\hat{j}\)
ⓑ. \(8\,\text{m s}^{-1}\hat{i}+3\,\text{m s}^{-1}\hat{j}\)
ⓒ. \(8\,\text{m s}^{-1}\hat{i}+7\,\text{m s}^{-1}\hat{j}\)
ⓓ. \(2\,\text{m s}^{-1}\hat{i}+3\,\text{m s}^{-1}\hat{j}\)
Correct Answer: \(8\,\text{m s}^{-1}\hat{i}+3\,\text{m s}^{-1}\hat{j}\)
Explanation: \( \textbf{Given position vector:} \) \(\vec{r}(t)=2t^2\,\hat{i}+(3t+1)\hat{j}\).
\( \textbf{Required:} \) Instantaneous velocity at \(t=2\,\text{s}\).
\( \textbf{Velocity relation:} \)
\[
\vec{v}=\frac{d\vec{r}}{dt}
\]
\( \textbf{Differentiate the \(x\)-component:} \)
\[
v_x=\frac{d}{dt}(2t^2)=4t
\]
\( \textbf{Differentiate the \(y\)-component:} \)
\[
v_y=\frac{d}{dt}(3t+1)=3
\]
\( \textbf{Velocity vector:} \)
\[
\vec{v}(t)=4t\,\hat{i}+3\hat{j}
\]
\( \textbf{Substitute \(t=2\,\text{s}\):} \)
\[
\vec{v}(2)=8\,\text{m s}^{-1}\hat{i}+3\,\text{m s}^{-1}\hat{j}
\]
\( \textbf{Final answer:} \) The instantaneous velocity is \(8\,\text{m s}^{-1}\hat{i}+3\,\text{m s}^{-1}\hat{j}\).
206. If the instantaneous velocity of a body is \(\vec{v}=6\,\text{m s}^{-1}\hat{i}-8\,\text{m s}^{-1}\hat{j}\), its instantaneous speed is:
ⓐ. \(2\,\text{m s}^{-1}\)
ⓑ. \(10\,\text{m s}^{-1}\)
ⓒ. \(14\,\text{m s}^{-1}\)
ⓓ. \(48\,\text{m s}^{-1}\)
Correct Answer: \(10\,\text{m s}^{-1}\)
Explanation: \( \textbf{Given velocity:} \) \(\vec{v}=6\,\text{m s}^{-1}\hat{i}-8\,\text{m s}^{-1}\hat{j}\).
\( \textbf{Required:} \) Instantaneous speed \(v=|\vec{v}|\).
\( \textbf{Components:} \) \(v_x=6\,\text{m s}^{-1}\) and \(v_y=-8\,\text{m s}^{-1}\).
\( \textbf{Speed relation:} \)
\[
v=\sqrt{v_x^2+v_y^2}
\]
\( \textbf{Substitution:} \)
\[
v=\sqrt{(6\,\text{m s}^{-1})^2+(-8\,\text{m s}^{-1})^2}
\]
\( \textbf{Squares:} \)
\[
(6)^2=36,\qquad (-8)^2=64
\]
\( \textbf{Calculation:} \)
\[
v=\sqrt{36+64}\,\text{m s}^{-1}=10\,\text{m s}^{-1}
\]
\( \textbf{Final answer:} \) The instantaneous speed is \(10\,\text{m s}^{-1}\).
207. Use the graph description below.
A body moves along a smooth curve in the \(x\)-\(y\) plane. At point \(P\), the tangent to the curve points toward the northeast.
The direction of the instantaneous velocity at \(P\) is:
ⓐ. toward the northeast
ⓑ. toward the centre of the coordinate system
ⓒ. along the straight line from the initial point to \(P\)
ⓓ. opposite to the tangent at \(P\)
Correct Answer: toward the northeast
Explanation: Instantaneous velocity at a point on a path is directed along the tangent to the path at that point. The graph description states that the tangent at \(P\) points toward the northeast. Therefore, the instantaneous velocity at \(P\) has the same direction. A line from the starting point to \(P\) may represent a displacement direction over a longer interval, not the instantaneous direction. The coordinate origin does not decide velocity direction unless the path’s tangent happens to point that way.
208. A table gives position vectors and claimed velocity vectors.
| Row | Position vector \(\vec{r}(t)\) | Claimed \(\vec{v}(t)\) |
|---|
| P | \(3t\,\hat{i}+2t^2\,\hat{j}\) | \(3\hat{i}+4t\,\hat{j}\) |
| Q | \((t^2+1)\hat{i}+5t\,\hat{j}\) | \(2t\,\hat{i}+5\hat{j}\) |
| R | \(4\hat{i}+t^3\,\hat{j}\) | \(4\hat{i}+3t^2\,\hat{j}\) |
| S | \(-2t\,\hat{i}+7\hat{j}\) | \(-2\hat{i}+0\hat{j}\) |
The row that contains a differentiation error is:
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row R
Explanation: Instantaneous velocity is obtained by differentiating the position vector with respect to time. Row P is correct because \(\frac{d}{dt}(3t)=3\) and \(\frac{d}{dt}(2t^2)=4t\). Row Q is correct because \(\frac{d}{dt}(t^2+1)=2t\) and \(\frac{d}{dt}(5t)=5\). Row S is correct because the derivative of the constant \(7\hat{j}\) is \(0\hat{j}\). Row R is wrong because the derivative of \(4\hat{i}\) is \(0\hat{i}\), not \(4\hat{i}\).
209. Three statements about instantaneous velocity are given.
I. It is written as \(\vec{v}=\frac{d\vec{r}}{dt}\).
II. Its magnitude is the instantaneous speed.
III. Its direction is always the same as the average velocity over any long interval.
The supported statements are:
ⓐ. I only
ⓑ. I and II only
ⓒ. II and III only
ⓓ. I, II, and III
Correct Answer: I and II only
Explanation: Statement I is correct because instantaneous velocity is the time derivative of position vector. Statement II is also correct because speed is the magnitude of velocity, so \(v=|\vec{v}|\). Statement III is not generally correct. Average velocity over a long interval points along the net displacement for that interval, while instantaneous velocity points along the tangent to the path at a particular point. These directions may be different for curved or non-uniform motion.
210. A moving body has the same average velocity over a time interval as another body. It does not necessarily mean that their instantaneous velocities at every instant are the same because:
ⓐ. average velocity only gives net displacement per unit time over the interval
ⓑ. instantaneous velocity has no direction
ⓒ. average velocity is always larger than instantaneous velocity
ⓓ. instantaneous velocity is calculated from total distance
Correct Answer: average velocity only gives net displacement per unit time over the interval
Explanation: Average velocity describes the net effect over a finite interval. It does not show how the velocity may have changed within that interval. Two bodies can have the same displacement in the same time but follow different paths or have different moment-to-moment velocities. Instantaneous velocity is local to one instant and is found from \(\vec{v}=\frac{d\vec{r}}{dt}\). The distinction is especially important in plane motion, where direction may change continuously.
211. The position vector of a body is \(\vec{r}(t)=(t^3-4t)\hat{i}+3t^2\hat{j}\), where \(\vec{r}\) is in \(\text{m}\) and \(t\) is in \(\text{s}\). Its speed at \(t=2\,\text{s}\) is:
ⓐ. \(4\sqrt{13}\,\text{m s}^{-1}\)
ⓑ. \(2\sqrt{13}\,\text{m s}^{-1}\)
ⓒ. \(4\sqrt{5}\,\text{m s}^{-1}\)
ⓓ. \(8\sqrt{13}\,\text{m s}^{-1}\)
Correct Answer: \(4\sqrt{13}\,\text{m s}^{-1}\)
Explanation: \( \textbf{Given position vector:} \) \(\vec{r}(t)=(t^3-4t)\hat{i}+3t^2\hat{j}\).
\( \textbf{Velocity relation:} \)
\[
\vec{v}=\frac{d\vec{r}}{dt}
\]
\( \textbf{x-component of velocity:} \)
\[
v_x=\frac{d}{dt}(t^3-4t)=3t^2-4
\]
\( \textbf{y-component of velocity:} \)
\[
v_y=\frac{d}{dt}(3t^2)=6t
\]
\( \textbf{At \(t=2\,\text{s}\):} \)
\[
v_x=3(2)^2-4=8\,\text{m s}^{-1}
\]
\[
v_y=6(2)=12\,\text{m s}^{-1}
\]
\( \textbf{Speed relation:} \)
\[
v=\sqrt{v_x^2+v_y^2}
\]
\( \textbf{Substitution:} \)
\[
v=\sqrt{8^2+12^2}\,\text{m s}^{-1}
\]
\( \textbf{Simplification:} \)
\[
v=\sqrt{208}\,\text{m s}^{-1}=4\sqrt{13}\,\text{m s}^{-1}
\]
\( \textbf{Final answer:} \) The speed at \(t=2\,\text{s}\) is \(4\sqrt{13}\,\text{m s}^{-1}\).
212. For \(\vec{r}(t)=(3t^2-12t)\hat{i}+4t\hat{j}\), a learner says the body is at rest at \(t=2\,\text{s}\) because the \(x\)-component of velocity becomes zero then. The better conclusion is:
ⓐ. the body is at rest because \(v_x=0\)
ⓑ. the body is at rest because position has an \(i\)-component
ⓒ. the body is not at rest because \(v_y\neq0\)
ⓓ. the body is not moving only if \(t=0\,\text{s}\)
Correct Answer: the body is not at rest because \(v_y\neq0\)
Explanation: \( \textbf{Position vector:} \) \(\vec{r}(t)=(3t^2-12t)\hat{i}+4t\hat{j}\).
\( \textbf{Velocity components:} \)
\[
v_x=\frac{d}{dt}(3t^2-12t)=6t-12
\]
\[
v_y=\frac{d}{dt}(4t)=4
\]
\( \textbf{At \(t=2\,\text{s}\):} \)
\[
v_x=6(2)-12=0
\]
\[
v_y=4\,\text{m s}^{-1}
\]
\( \textbf{Rest condition:} \) A body is at rest only when all velocity components are zero.
\( \textbf{Velocity at \(t=2\,\text{s}\):} \)
\[
\vec{v}=0\hat{i}+4\,\text{m s}^{-1}\hat{j}
\]
\( \textbf{Final answer:} \) The body is not at rest because it still has a non-zero \(y\)-component of velocity.
213. The position vector of a body is \(\vec{r}(t)=(t^2-4t)\hat{i}+5t\hat{j}\). At what time is its velocity perpendicular to the \(x\)-axis?
ⓐ. \(1\,\text{s}\)
ⓑ. \(4\,\text{s}\)
ⓒ. \(5\,\text{s}\)
ⓓ. \(2\,\text{s}\)
Correct Answer: \(2\,\text{s}\)
Explanation: \( \textbf{Given:} \) \(\vec{r}(t)=(t^2-4t)\hat{i}+5t\hat{j}\).
\( \textbf{Velocity relation:} \)
\[
\vec{v}=\frac{d\vec{r}}{dt}
\]
\( \textbf{x-component of velocity:} \)
\[
v_x=\frac{d}{dt}(t^2-4t)=2t-4
\]
\( \textbf{y-component of velocity:} \)
\[
v_y=\frac{d}{dt}(5t)=5
\]
\( \textbf{Condition for velocity perpendicular to \(x\)-axis:} \) The velocity must have \(v_x=0\).
\( \textbf{Apply condition:} \)
\[
2t-4=0
\]
\( \textbf{Solve:} \)
\[
t=2\,\text{s}
\]
\( \textbf{Direction check:} \) At this instant, \(v_y=5\,\text{m s}^{-1}\), so the velocity is along the \(y\)-direction.
\( \textbf{Final answer:} \) The velocity is perpendicular to the \(x\)-axis at \(t=2\,\text{s}\).
214. Use the graph information below.
At a certain instant, the slope of the \(x-t\) graph is \(3\,\text{m s}^{-1}\), and the slope of the \(y-t\) graph is \(-4\,\text{m s}^{-1}\).
The instantaneous speed at that instant is:
ⓐ. \(1\,\text{m s}^{-1}\)
ⓑ. \(5\,\text{m s}^{-1}\)
ⓒ. \(7\,\text{m s}^{-1}\)
ⓓ. \(12\,\text{m s}^{-1}\)
Correct Answer: \(5\,\text{m s}^{-1}\)
Explanation: \( \textbf{Graph meaning:} \) The slope of the \(x-t\) graph gives \(v_x\).
\( \textbf{x-component:} \)
\[
v_x=3\,\text{m s}^{-1}
\]
\( \textbf{Graph meaning for \(y\):} \) The slope of the \(y-t\) graph gives \(v_y\).
\( \textbf{y-component:} \)
\[
v_y=-4\,\text{m s}^{-1}
\]
\( \textbf{Speed relation:} \)
\[
v=\sqrt{v_x^2+v_y^2}
\]
\( \textbf{Substitution:} \)
\[
v=\sqrt{(3\,\text{m s}^{-1})^2+(-4\,\text{m s}^{-1})^2}
\]
\( \textbf{Calculation:} \)
\[
v=\sqrt{9+16}\,\text{m s}^{-1}=5\,\text{m s}^{-1}
\]
\( \textbf{Final answer:} \) The instantaneous speed is \(5\,\text{m s}^{-1}\).
215. Assertion: Instantaneous velocity at a point on a curved path is along the tangent at that point.
Reason: In the limiting process, the displacement over a very small time interval approaches the tangent direction.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The assertion is true because instantaneous velocity gives the direction of motion at one point of the path. For a curved path, average velocity over a finite interval points along a chord joining two nearby points. As the time interval becomes smaller, the second point comes closer to the first point. The chord then approaches the tangent at that point. The reason describes exactly why the limiting direction of velocity is tangent to the path.
216. A path is described as follows: near point \(P\), the body’s curve is momentarily horizontal, and the body is moving from left to right. The instantaneous velocity at \(P\) is:
ⓐ. vertically upward
ⓑ. vertically downward
ⓒ. horizontal toward the right
ⓓ. opposite to the direction of motion
Correct Answer: horizontal toward the right
Explanation: Instantaneous velocity follows the tangent direction at the point of motion. The path is momentarily horizontal near \(P\), so the tangent at \(P\) is horizontal. The body is moving from left to right, which fixes the sense of the velocity along that tangent. A horizontal tangent does not by itself decide left or right; the motion sense supplies that information. The velocity is therefore horizontal toward the right.
217. Average acceleration in a plane is defined by:
ⓐ. \(\vec{a}_{\text{avg}}=\frac{\Delta\vec{r}}{\Delta t}\)
ⓑ. \(\vec{a}_{\text{avg}}=\vec{v}\Delta t\)
ⓒ. \(\vec{a}_{\text{avg}}=\frac{\text{total distance}}{\Delta t}\)
ⓓ. \(\vec{a}_{\text{avg}}=\frac{\Delta\vec{v}}{\Delta t}\)
Correct Answer: \(\vec{a}_{\text{avg}}=\frac{\Delta\vec{v}}{\Delta t}\)
Explanation: Average acceleration is the rate of change of velocity over a time interval. Since velocity is a vector, its change \(\Delta\vec{v}\) is also a vector. Dividing this vector by the positive time interval \(\Delta t\) gives \(\vec{a}_{\text{avg}}=\frac{\Delta\vec{v}}{\Delta t}\). The expression \(\frac{\Delta\vec{r}}{\Delta t}\) gives average velocity, not average acceleration. The unit of average acceleration is \(\text{m s}^{-2}\), reflecting change in velocity per unit time.
218. A body can have non-zero acceleration even when its speed is constant if:
ⓐ. its velocity vector remains completely constant
ⓑ. its displacement is zero at every instant
ⓒ. the direction of its velocity changes
ⓓ. its time interval is not measured
Correct Answer: the direction of its velocity changes
Explanation: Acceleration depends on change in velocity, not only on change in speed. Velocity is a vector, so a change in direction is also a change in velocity. If the speed remains constant but the direction of motion changes, the velocity vector changes. Therefore, acceleration can be non-zero. This is a common feature of curved motion, where the tangent direction of velocity changes from point to point.
219. The velocity of a body changes from \(\vec{v}_1=3\,\text{m s}^{-1}\hat{i}+4\,\text{m s}^{-1}\hat{j}\) to \(\vec{v}_2=-5\,\text{m s}^{-1}\hat{i}+10\,\text{m s}^{-1}\hat{j}\) in \(2\,\text{s}\). Its average acceleration is:
ⓐ. \(-4\,\text{m s}^{-2}\hat{i}+3\,\text{m s}^{-2}\hat{j}\)
ⓑ. \(4\,\text{m s}^{-2}\hat{i}-3\,\text{m s}^{-2}\hat{j}\)
ⓒ. \(-8\,\text{m s}^{-2}\hat{i}+6\,\text{m s}^{-2}\hat{j}\)
ⓓ. \(-1\,\text{m s}^{-2}\hat{i}+7\,\text{m s}^{-2}\hat{j}\)
Correct Answer: \(-4\,\text{m s}^{-2}\hat{i}+3\,\text{m s}^{-2}\hat{j}\)
Explanation: \( \textbf{Initial velocity:} \) \(\vec{v}_1=3\,\text{m s}^{-1}\hat{i}+4\,\text{m s}^{-1}\hat{j}\).
\( \textbf{Final velocity:} \) \(\vec{v}_2=-5\,\text{m s}^{-1}\hat{i}+10\,\text{m s}^{-1}\hat{j}\).
\( \textbf{Time interval:} \) \(\Delta t=2\,\text{s}\).
\( \textbf{Velocity change:} \)
\[
\Delta\vec{v}=\vec{v}_2-\vec{v}_1
\]
\( \textbf{x-component change:} \)
\[
\Delta v_x=-5\,\text{m s}^{-1}-3\,\text{m s}^{-1}=-8\,\text{m s}^{-1}
\]
\( \textbf{y-component change:} \)
\[
\Delta v_y=10\,\text{m s}^{-1}-4\,\text{m s}^{-1}=6\,\text{m s}^{-1}
\]
\( \textbf{Average acceleration:} \)
\[
\vec{a}_{\text{avg}}=\frac{\Delta\vec{v}}{\Delta t}
\]
\( \textbf{Component division:} \)
\[
a_{\text{avg},x}=\frac{-8\,\text{m s}^{-1}}{2\,\text{s}}=-4\,\text{m s}^{-2}
\]
\[
a_{\text{avg},y}=\frac{6\,\text{m s}^{-1}}{2\,\text{s}}=3\,\text{m s}^{-2}
\]
\( \textbf{Final answer:} \) \(\vec{a}_{\text{avg}}=-4\,\text{m s}^{-2}\hat{i}+3\,\text{m s}^{-2}\hat{j}\).
220. A table compares velocity-change records.
| Row | Velocity change \(\Delta\vec{v}\) | Time \(\Delta t\) | Claimed \(\vec{a}_{\text{avg}}\) |
|---|
| P | \(6\,\text{m s}^{-1}\hat{i}\) | \(3\,\text{s}\) | \(2\,\text{m s}^{-2}\hat{i}\) |
| Q | \(-8\,\text{m s}^{-1}\hat{j}\) | \(4\,\text{s}\) | \(-2\,\text{m s}^{-2}\hat{j}\) |
| R | \(5\,\text{m s}^{-1}\hat{i}+10\,\text{m s}^{-1}\hat{j}\) | \(5\,\text{s}\) | \(\text{m s}^{-2}\hat{i}+2\,\text{m s}^{-2}\hat{j}\) |
| S | \(4\,\text{m s}^{-1}\hat{i}-6\,\text{m s}^{-1}\hat{j}\) | \(2\,\text{s}\) | \(2\,\text{m s}^{-2}\hat{i}+3\,\text{m s}^{-2}\hat{j}\) |
The row that contains a sign error is:
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row S
Explanation: Average acceleration is found from \(\vec{a}_{\text{avg}}=\frac{\Delta\vec{v}}{\Delta t}\). Row P is correct because \(\frac{6\,\text{m s}^{-1}}{3\,\text{s}}=2\,\text{m s}^{-2}\). Row Q is correct because \(\frac{-8\,\text{m s}^{-1}}{4\,\text{s}}=-2\,\text{m s}^{-2}\). Row R is correct because each component is divided by \(5\,\text{s}\), giving \(1\,\text{m s}^{-2}\hat{i}+2\,\text{m s}^{-2}\hat{j}\). Row S is wrong because \(\frac{-6\,\text{m s}^{-1}}{2\,\text{s}}=-3\,\text{m s}^{-2}\), not \(+3\,\text{m s}^{-2}\).
