4 The position-time graph in Figure 12 shows the motion an object that has been thrown vertically upwards with a velocity of $$ 12 \mathrm{~m} . \mathrm{s}^{-1} $$. 4.1 Write down its velocity (magnitude and direction) at time 4.1.1 $$ t_{1} $$ 4.1.2 $$ \quad t_{2} $$ 4.1.3 $$ \quad t_{3} $$ 4.2 Draw the following rough sketches of the object's motion 4.2.1 velocity-time graph taking upwards as the positive direction. 4.2.2 velocity-time graph taking downwards as the positi direction. 4.2.3 acceleration-time graph taking upwards as the posi direction. (Remember that the object's acceleration constant, g.) Answers to numerical questions 2.3.1 $$ 6,13 \mathrm{~m} $$ above the player's hand; $$ 366,15 \mathrm{~m} $$ to $$ 66,29 \mathrm{~m} $$ (depending on roundir

Question

UpStudy AI · Accepted Answer

To solve the problem, we need to analyze the motion of an object thrown vertically upwards with an initial velocity of $$ 12 \, \text{m/s} $$. We will extract the known conditions and then address each part of the question step by step.

### Known Conditions:
- Initial velocity $$ v_0 = 12 \, \text{m/s} $$ (upwards)
- Acceleration due to gravity $$ g = 9.81 \, \text{m/s}^2 $$ (downwards)

### 4.1 Velocity at Different Times
To find the velocity at different times $$ t_1, t_2, $$ and $$ t_3 $$, we can use the formula for velocity under constant acceleration:

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

#### 4.1.1 Velocity at $$ t_1 $$
Assuming $$ t_1 $$ is the time when the object is still moving upwards but has slowed down, we need to know the specific time value to calculate the velocity. However, if we assume $$ t_1 $$ is before the peak, we can express it as:

$$
v(t_1) = 12 - 9.81 \cdot t_1
$$

#### 4.1.2 Velocity at $$ t_2 $$
Assuming $$ t_2 $$ is the time at the peak of the motion, where the velocity becomes zero:

$$
v(t_2) = 0
$$

This occurs when:

$$
0 = 12 - 9.81 \cdot t_2 \implies t_2 = \frac{12}{9.81} \approx 1.22 \, \text{s}
$$

#### 4.1.3 Velocity at $$ t_3 $$
Assuming $$ t_3 $$ is after the peak and the object is falling back down, we can express it as:

$$
v(t_3) = 12 - 9.81 \cdot t_3
$$

### 4.2 Sketches of the Object's Motion
#### 4.2.1 Velocity-Time Graph (Upwards as Positive)
- The graph starts at $$ 12 \, \text{m/s} $$ and decreases linearly to $$ 0 \, \text{m/s} $$ at $$ t_2 $$.
- After $$ t_2 $$, the velocity becomes negative as the object falls back down.

#### 4.2.2 Velocity-Time Graph (Downwards as Positive)
- The graph starts at $$ -12 \, \text{m/s} $$ (downwards) and increases to $$ 0 \, \text{m/s} $$ at $$ t_2 $$.
- After $$ t_2 $$, the velocity continues to increase positively as the object falls.

#### 4.2.3 Acceleration-Time Graph (Upwards as Positive)
- The acceleration is constant and equal to $$ -9.81 \, \text{m/s}^2 $$ (downwards), so the graph is a horizontal line at $$ -9.81 $$.

### Summary of Answers
- **4.1.1**: $$ v(t_1) = 12 - 9.81 \cdot t_1 $$ (upwards)
- **4.1.2**: $$ v(t_2) = 0 $$ (at the peak)
- **4.1.3**: $$ v(t_3) = 12 - 9.81 \cdot t_3 $$ (downwards)

For the sketches, you would typically draw the graphs based on the descriptions provided. If you need further assistance with the sketches or any specific calculations, please provide the time values for $$ t_1 $$ and $$ t_3 $$.
At time $$ t_1 $$, the velocity is $$ 12 - 9.81 \times t_1 $$ m/s upwards. At $$ t_2 $$, the velocity is 0 m/s at the peak. At $$ t_3 $$, the velocity is $$ 12 - 9.81 \times t_3 $$ m/s downwards. For the velocity-time graph with upwards as positive, it starts at 12 m/s and decreases to 0 m/s at $$ t_2 $$, then becomes negative as the object falls. For the velocity-time graph with downwards as positive, it starts at -12 m/s and increases to 0 m/s at $$ t_2 $$, then becomes positive as the object falls. The acceleration-time graph is a horizontal line at -9.81 m/s².

Upstudy AI Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Physics Questions