University

Physics 1

Physics 1

Fall 2024

- Concept Questions
- c1. Motion Graphs
- c2. Velocity from Graph
- c3. Graphs from motion description
- c4. Velocity and acceleration graphs from position graph
- c5. Free-fall velocity and acceleration
- c6. Interpreting a position graph
- c7. Velocity from acceleration
- c8. Sign of acceleration
- c9. Motion of an accelerating cart
- c10. Units of acceleration

- Long Answer Questions

An object is thrown upward and caught when it returns to the same place. If up is the positive direction, sketch the position, velocity and acceleration graphs for this motion.

The graph shows an object’s position versus time.

- What is the object’s average velocity from time \(t_0=1\) s to \(t_1=4\) s?
- What is the object’s average velocity from time \(t_0=1\) s to \(t_1=7\) s?
- What is the object’s average velocity from time \(t_0=2\) s to \(t_1=7\) s?
- What is the object’s instantaneous velocity at time \(t=4\) s?
- What is the object’s approxmiate instantaneous velocity at time \(t=2\) s?

\(1.33\) m/s

\(0\) m/s

\(-0.4\) m/s

\(0\) m/s

about \(2\) m/s, estimating the slope

Consider a man walking on an \(x\) axis, where positive \(x\) points to the right. The man leaves the origin, speeding up while moving to the right. He then begins to slow down and then reverses direction and speeds up moving back toward the origin. Shortly before passing the origin he is moving at a constant speed to the left. He passes the origin and continues moving at a constant speed to the left.

Sketch the position, velocity and acceleration graphs for this motion.

The graph shows the position of an object subject to a constant acceleration. Sketch the velocity and acceleration graphs.

A ball is tossed upward (\(+\) direction) and is in free fall. When the ball is at its highest point, give the signs (\(+\), \(-\) or \(0\)) of the velocity and acceleration of the ball.

Velocity is \(0\). Acceleration is \(-\).

The figure shows a position-time graph for a moving object. The \(x\) axis points to the right. Give the letter(s) for the following:

- Object is moving fastest.
- Object is moving left.
- Object is speeding up.
- Object is slowing down.
- Object is turning around.

- D
- C, D, E
- C
- A, E
- B

Below are three acceleration-time graphs. For each, draw the corresponding velocity-time graph. You may assume that the object starts at rest.

Plus and minus signs are used in one-dimensional motion to indicate direction. What is the sign of an acceleration that reduces the magnitude of a negative velocity? Of a positive velocity?

Plus, minus

A cart speeds up toward the origin. What do the position and velocity graphs look like?

Which of the following could be units of acceleration?

- mi/hr
- mi/hr/s
- m²/s²
- m·s\(^{-2}\)
- mi²/hr/s
- \(\frac{\rm{m/s}}{\rm{s}}\)

b, d, f

These are the only ones with dimension \(L/T^2\).

A hot-air balloon rises from ground level at a constant velocity of 3.0 m/s. One minute after liftoff, a sandbag is dropped accidentally from the balloon. Calculate (a) the time it takes for the sandbag to reach the ground and (b) the speed of the sandbag when it hits the ground.

\(t = 6.37\) s, taking the positive root

\(v = 59.5\) m/s

An object is dropped (from rest) from a height of \(75.0\) m above ground level.

- Determine the distance traveled during the first second.
- Determine the final speed at which the object hits the ground.
- Determine the distance traveled during the last second of motion before hitting the ground.

\(\Delta y = v_0 t - \tfrac{1}{2} g t^2 = - \tfrac{1}{2} g t^2\)

\(v^2 = v_0^2 - 2 g h = - 2 g h\)

\(v = v_0 - g t = - g t\)

\(y = 4.9\) m

\(v = 38.3\) m/s

\(33.4\) m

A particle travels along a straight line with a velocity of \(v = (4t - 3t^2)\) m/s, where \(t\) is in seconds. Determine the position of the particle when \(t=4\) s, given that \(s = 0\) when \(t = 0\).

Displacement is given by \(\displaystyle \int_{t_0}^{t} v(t)\; dt = s(t) - s_0\).

Initial conditions are \(t_0 = 0\) and \(s_0 = 0\). So

\[s(t) = \int_{0}^{t} (4t - 3t^2) \; dt = ½ \cdot 4 t^2 - ⅓ \cdot 3 t^3 = 2 t^2 - t^3 \]

At \(t=4\) s,

\[s = -32 \; \rm{m}\]

A particle travels along a straight line with an acceleration of \(a = (10 - 0.2s)\) m/s², where \(s\) is measured in meters. Determine the velocity of the particle when \(s=10\) m if \(v=5\) m/s at \(s=0\).

Acceleration is given by \(\displaystyle \int_{s_0}^{s} a(s)\; ds = \tfrac{1}{2} v^2 - \tfrac{1}{2} v_0^2\)

Initial conditions are \(v_0 = 5\) m/s and \(s_0 = 0\). So

\[ \tfrac{1}{2} v^2(s) - \tfrac{1}{2} \cdot 25 = \int_{0}^{s} (10 - 0.2s) \; ds = 10s - ½ \cdot 0.2 s^2 = 10s - 0.1 s^2 \]

Solve this for \(v(s)\):

\[v(s) = \sqrt{ 20s - 0.2 s^2 + 25 }\]

At \(s=10\) m,

\[v = 14.3 \; \rm{m/s}\]

*Last modified: Mon September 23 2024, 08:39 AM.*