The velocity–time graph for an object is shown.
Which expression gives the total displacement of the object?
Correct Answer: \( \boxed{\mathrm{A}} \)
Displacement is the signed area under a velocity–time graph.
Area \(1\) is above the time axis and is positive, while area \(2\) is below the time axis and is negative.
Hence, the total displacement is \( \text{area }1-\text{area }2 \).
Therefore, the correct answer is (A).
The graph shows the variation of velocity with time \(t\) of an object moving in a straight line.
At \(t=0\), the displacement of the object is zero.
What is the displacement of the object at \(t=20\,\mathrm{s}\)?
Correct Answer: \( \boxed{\mathrm{B}} \)
Displacement is the area under the velocity-time graph.
From \(0\) to \(8\,\mathrm{s}\):
\(A_1=\dfrac{1}{2}\times8\times3=12\,\mathrm{m}\).
From \(8\) to \(16\,\mathrm{s}\):
\(A_2=\dfrac{1}{2}\times8\times3=12\,\mathrm{m}\).
From \(16\) to \(20\,\mathrm{s}\):
\(A_3=-\dfrac{1}{2}\times4\times1.5=-3\,\mathrm{m}\).
Total displacement:
\(12+12-3=21\,\mathrm{m}\).
Therefore, the correct answer is (B).
An astronaut on the Moon, where there is no air resistance, throws a ball. The ball’s initial velocity has a vertical component of \(8.00\,\mathrm{m\,s^{-1}}\) and a horizontal component of \(4.00\,\mathrm{m\,s^{-1}}\), as shown.
The acceleration of free fall on the Moon is \(1.62\,\mathrm{m\,s^{-2}}\).
What is the speed of the ball \(9.00\,\mathrm{s}\) after being thrown?
Correct Answer: \( \boxed{\mathrm{B}} \)
The horizontal velocity remains constant because there is no air resistance.
\(v_x=4.00\,\mathrm{m\,s^{-1}}\)
The vertical velocity after \(9.00\,\mathrm{s}\) is
\(v_y=u_y-gt=8.00-(1.62)(9.00)=-6.58\,\mathrm{m\,s^{-1}}\)
The speed is the magnitude of the velocity:
\(v=\sqrt{v_x^2+v_y^2}=\sqrt{4.00^2+(-6.58)^2}=\sqrt{59.30}=7.70\,\mathrm{m\,s^{-1}}\)
Therefore, the correct answer is (B).
