Unit: 1. Kinematics
Weightage : 10-15%
Two people are in a boat that is capable of a maximum speed of \(5\,\mathrm{km/h}\) in still water and wish to cross a river \(1\,\mathrm{km}\) wide to a point directly across from their starting point. If the speed of the water in the river is \(5\,\mathrm{km/h}\), how much time is required for the crossing?
Correct Answer: \( \boxed{\mathrm{E}} \)
To land directly opposite the starting point, the boat’s upstream velocity component must exactly cancel the river’s current.
The river current is
\(v_{\mathrm{river}}=5\,\mathrm{km/h}\)
The boat’s maximum speed relative to the water is also
\(v_{\mathrm{boat}}=5\,\mathrm{km/h}\)
Therefore, the boat must point directly upstream so that its entire speed cancels the current:
\(v_{\mathrm{boat,\,upstream}}=v_{\mathrm{river}}=5\,\mathrm{km/h}\)
This leaves no component of the boat’s velocity directed across the river:
\(v_{\perp}=0\)
Since there is no velocity across the river, the boat can never reach the opposite bank directly across from its starting point.
Therefore, the destination cannot be reached under these conditions.
Hence, the correct answer is (E).
A train is traveling at a constant speed eastward, and the velocity vector \(v_T\) is shown in the diagram above. The velocity vector \(v_C\) of a cyclist moving at a direction of \(15.0^\circ\) east of north is also shown. Which of the following vectors could represent the velocity of the cyclist as seen by someone on the train?
Correct Answer: \( \boxed{\mathrm{B}} \)
The velocity of the cyclist relative to the train is found using relative velocity:
\( \vec{v}_{C/T}=\vec{v}_C-\vec{v}_T \).
Since the train moves east, subtracting \( \vec{v}_T \) is equivalent to adding a vector of the same magnitude pointing west.
The cyclist’s velocity is directed \(15.0^\circ\) east of north, so it has a large northward component and a smaller eastward component. After subtracting the train’s eastward velocity, the eastward component decreases and can become westward, while the northward component remains unchanged.
Therefore, the cyclist appears to move upward and to the left (northwest) relative to the train.
Among the given choices, only Option (B) shows a vector pointing up and to the left, so it correctly represents the cyclist’s velocity as observed from the train.
Student 1 is standing on a cart holding a small stone, while student 2 is standing at rest on the ground, as shown in the figure above. The cart is moving at a constant speed \(v\) in the \(+x\)-direction, as indicated by the coordinate system shown. Student 1 drops the stone precisely when passing student 2.
Which of the following best represents the path of the falling stone relative to student 1 and the path of the falling stone relative to student 2?
Correct Answer: \( \boxed{\mathrm{C}} \)
Analyze the motion in the two different reference frames.
Relative to student 1 (moving with the cart): At the instant the stone is released, it has zero horizontal velocity relative to the cart. Since the cart continues moving at constant speed and there is no horizontal acceleration in this frame, the stone falls straight downward.
Relative to student 2 (standing on the ground): The stone has an initial horizontal velocity \(v\) equal to the speed of the cart. Gravity provides only a vertical acceleration,
\(\vec{a}=(0,-g)\)
Therefore, the horizontal velocity remains constant while the vertical speed increases, producing a projectile trajectory described by
\(x=vt,\qquad y=h-\dfrac{1}{2}gt^2\)
Thus, the stone appears to fall straight down to student 1 but follows a parabolic path curving forward relative to student 2.
Hence, the correct answer is (C).
