Unit 3: Linear Motion

Zero.

Constant velocity means that the magnitude and direction of the velocity vector do not change with time. So, during the 10-s time interval, the car experiences a change in speed of zero. Therefore, the acceleration of the car is equal to \begin{equation} \frac{\Delta v}{\Delta t} = \frac{0}{10\,\text{s}} = 0\end{equation}

Acceleration increases as the steepness of the incline increases. When the plane is vertical, the ball is in free fall and its acceleration is $g$.

An initially stationary object on a horizontal surface will remain at rest if not disturbed, and its acceleration is equal to zero. Galileo found that on an inclined plane, the steeper the incline, the greater the acceleration of an object placed on it. When the incline is angled vertically at 90°, the acceleration reaches its maximum free-fall value of $g = 10\,\text{m}/\text{s}^2\textrm{.}$

2.5 m/s²

Assuming a steady increase in the car’s speed, the acceleration is calculated as follows: \begin{align} a &= \frac{\text{change in speed}}{\text{time interval}} \nonumber\\[6pt] &= \frac{\Delta v}{t} \nonumber\\[6pt] &= \frac{25\,\text{m/s} - 0}{10\,\text{s}} \nonumber\\[6pt] &= 2.5\,\text{m}/\text{s}^2 \end{align} You can then calculate the distance traveled by the car as follows: \begin{align} d &= \tfrac{1}{2} at^2 \nonumber\\[6pt] &= \tfrac{1}{2}\, (2.5\,\text{m}/\text{s}^2)\, (10\,\text{s})^2 \nonumber\\[6pt] &= 125\,\text{m} \end{align}

45 m, 6 s

By ignoring air resistance, you can treat the entire motion as free fall, where the ball experiences a downward acceleration of $g = 10\,\text{m}/\text{s}^2$ while it is in the air. During the ascending part of its flight, the ball slows down uniformly from an initial speed of 30 m/s to a final speed of zero at the maximum height $h$. So the duration ($t_1$) of the first half of the ball’s trip is calculated as follows: \begin{align} \Delta v &= gt_1 \nonumber\\[6pt] \Rightarrow\quad t_1 &= \frac{\Delta v}{g} \nonumber\\[6pt] &= \frac{30\,\text{m/s}}{10\,\text{m}/\text{s}^2} \nonumber\\[6pt] &= 3\,\text{s} \end{align}

After it stops momentarily at the maximum height, the ball begins the descending part of its flight, where its downward speed increases by 10 m/s each second. Due to the symmetry of both parts of its motion, the ball gains a final speed of 30 m/s, which is equal to the launch speed (see Video Q3.44.) Also, the time it takes the ball to rise to the maximum height is equal to the time it takes to fall back to your hand. If $t_\text{air}$ is the total time the ball is in the air, then \begin{equation} t_\text{air} = 2 t_1 = 6\,\text{s} \end{equation} Therefore, you can calculate the vertical distance traveled by the ball as it ascends as \begin{align} h &= \tfrac{1}{2} g\,t_1^2 \nonumber\\[6pt] &= \tfrac{1}{2} (10\,\text{m}/\text{s}^2) (3\,\text{s})^2 \nonumber\\[6pt] &= 45\,\text{m} \end{align}

Suzie’s canoe will not move relative to the riverbank.

The canoe travels upstream at a speed of 8 km/h relative to the water. However, the water carrying the canoe flows downstream at the same speed. So, with respect to a person standing on the riverbank, the speed of the canoe is equal to zero and the canoe is observed to remain in the same spot.

Both airplanes move at the same speed, but they have different velocities.

Speed is defined as the distance traveled by an object per unit time, regardless of the direction of motion. Both airplanes cover 300 km in one hour and therefore have the same speed. Velocity, however, is a vector quantity that incorporates direction. In this case, the first airplane has a velocity of 300 km/h due north, while the velocity of the other airplane is 300 km/h due south.

You do not have enough information to answer this question.

Acceleration is the rate at which the velocity of an object changes with time. For example, if the first car achieves the speed of 50 km/h in a period of 5 seconds, its acceleration is equal to \begin{equation} \frac{50\,\text{km/h}}{5\,\text{s}} = 10\,\text{km/h}\;\text{per second} \end{equation} However, if the other car takes 10 seconds to reach the speed of 60 km/h, then it accelerates at the slower rate of \begin{equation} \frac{60\,\text{km/h}}{10\,\text{s}} = 6\,\text{km/h}\;\text{per second} \end{equation} Since the acceleration periods are not given, you have no way of knowing which car underwent the greater acceleration.

In both cases, the ball experiences a downward acceleration of 10 m/s².

The free-fall acceleration does not depend on the mass, speed, or direction of motion of an object if it is moving under the influence of gravity only.

In the absence of air resistance, freely falling raindrops would reach very high speeds.

Assume a raindrop leaves a cloud at an altitude of $h = 2000\,\text{m}$. In the absence of air resistance, the time it would take the raindrop to reach the ground is calculated as follows: \begin{align} h &= \tfrac{1}{2} g t^2 \nonumber\\[6pt] \Rightarrow\quad t &= \sqrt{\frac{2h}{g}} \nonumber\\[6pt] &= \sqrt{\frac{2 \times 2000\,\text{m}}{10\,\text{m}/\text{s}^2}} \nonumber\\[6pt] &= 20\,\text{s} \end{align} So you can calculate the final speed of the raindrop as follows: \begin{align} v &= g t \nonumber\\[6pt] &= (10\,\text{m}/\text{s}^2) \, (20\,\text{s}) \nonumber\\[6pt] &= 200\,\text{m/s}\quad (\text{or}\; 720\,\text{km/h}) \end{align} This is a very high speed, comparable to that of shotgun pellet! Air resistance prevents rain drops from gaining such dangerous speeds.

Yes, when the automobile is slowing down.

Imagine driving your car north at a speed of 72 km/h, which corresponds to 20 m/s. When you see a deer on the road, you slam on the brakes and bring the car to a stop in 4 s. During this period, the car’s velocity changes from 20 m/s due north to zero. The change in the velocity vector can be expressed mathematically as follows: \begin{align} \Delta \bf{v} &= 0 - 20\,\text{m/s}\;\text{due north} \nonumber\\[6pt] &= - 20\,\text{m/s}\;\text{due north} \nonumber\\[6pt] &= 20\,\text{m/s}\;\text{due south} \end{align} Since acceleration is the change in velocity per unit time, you can calculate the average acceleration of the car while stopping as follows: \begin{align} \Delta {\bf a} &= \frac{\Delta {\bf v}}{\Delta t} \nonumber\\[6pt] &= \frac{20\,\text{m/s}\;\text{due south}}{4\,\text{s}} \nonumber\\[6pt] &= 5\,\text{m}/\text{s}^2\;\text{due south} \end{align}

Yes, it is possible.

When you throw a rock vertically upward, its speed decreases uniformly at a rate equal to the acceleration due to gravity. After it stops (momentarily) at maximum height, the rock starts descending and its downward speed increases at the same rate of 10 m/s every second. So, in this case, the rock reverses its direction of travel while its acceleration (10 m/s² downward) remains unchanged.

Think about the state of an object that is reversing its direction of motion.

Yes, it is possible.

When you throw a ball upward, it stops momentarily when it reaches its maximum height. At this moment, even though the speed of the ball is equal to zero, it continues to experience a downward acceleration of 10 m/s² due to gravity.