https://file.aiquickdraw.com/mathbot/file/1727323203846i3rh9osn.jpg

The image presents three physics problems involving linear motion (evaluasi 3.4):

1. **Problem 1**: A body moves in a straight line with a constant acceleration of 4 m/s². If the object starts from rest, determine the velocity and distance traveled by the object after moving for 10 seconds.

2. **Problem 2**: A car initially moves at a speed of 36 km/h. After moving for 10 seconds, the car's speed becomes 108 km/h. Determine the acceleration of the car and the distance it travels during that time.

3. **Problem 3**: Object A initially moves at a speed of 18 km/h. After traveling a distance of 200 meters, the speed of the object changes to 90 km/h. Calculate the time required for the object to cover that distance.

Let me solve these one by one.

---

### **Solution 1**:
Given:
- Acceleration, \( a = 4 \, \text{m/s}^2 \)
- Initial velocity, \( u = 0 \, \text{m/s} \)
- Time, \( t = 10 \, \text{seconds} \)

#### 1.1. Velocity after 10 seconds:
We can use the equation for velocity:
\[
v = u + at
\]
Substitute the values:
\[
v = 0 + (4 \times 10) = 40 \, \text{m/s}
\]
So, the velocity after 10 seconds is **40 m/s**.

#### 1.2. Distance traveled in 10 seconds:
We can use the equation for distance:
\[
s = ut + \frac{1}{2} a t^2
\]
Substitute the values:
\[
s = 0 + \frac{1}{2} (4) (10)^2 = 2 \times 100 = 200 \, \text{m}
\]
So, the distance traveled is **200 meters**.

---

### **Solution 2**:
Given:
- Initial speed, \( u = 36 \, \text{km/h} = 10 \, \text{m/s} \) (converted from km/h to m/s)
- Final speed, \( v = 108 \, \text{km/h} = 30 \, \text{m/s} \)
- Time, \( t = 10 \, \text{seconds} \)

#### 2.1. Acceleration:
We can use the equation for acceleration:
\[
a = \frac{v - u}{t}
\]
Substitute the values:
\[
a = \frac{30 - 10}{10} = \frac{20}{10} = 2 \, \text{m/s}^2
\]
So, the acceleration is **2 m/s²**.

#### 2.2. Distance traveled:
We can use the equation for distance:
\[
s = ut + \frac{1}{2} a t^2
\]
Substitute the values:
\[
s = (10 \times 10) + \frac{1}{2} (2) (10)^2 = 100 + 1 \times 100 = 200 \, \text{m}
\]
So, the distance traveled is **200 meters**.

---

### **Solution 3**:
Given:
- Initial speed, \( u = 18 \, \text{km/h} = 5 \, \text{m/s} \)
- Final speed, \( v = 90 \, \text{km/h} = 25 \, \text{m/s} \)
- Distance, \( s = 200 \, \text{m} \)

#### 3.1. Time required:
First, we need to find the acceleration. We can use the following equation:
\[
v^2 = u^2 + 2as
\]
Solving for \( a \):
\[
25^2 = 5^2 + 2a(200)
\]
\[
625 = 25 + 400a
\]
\[
600 = 400a
\]
\[
a = \frac{600}{400} = 1.5 \, \text{m/s}^2
\]
Now, using the equation \( v = u + at \), we can solve for time \( t \):
\[
25 = 5 + (1.5) t
\]
\[
25 - 5 = 1.5 t
\]
\[
t = \frac{20}{1.5} = 13.33 \, \text{seconds}
\]
So, the time required is **13.33 seconds**.

---

Would you like more details or have any questions about these solutions?

Here are 5 related questions you might want to explore:
1. How do we calculate the final velocity if the time and acceleration are unknown?
2. Can you explain the difference between uniform acceleration and variable acceleration?
3. What happens if the object in the first problem had an initial velocity?
4. How would the equations change if the motion occurred on an inclined plane?
5. How do we determine displacement when the object experiences non-uniform acceleration?

**Tip**: Always remember to convert units (like km/h to m/s) when solving physics problems to maintain consistency in calculations.

Physics problems involving motion with constant acceleration.

Explore how to solve physics problems using kinematic equations. This includes calculating final velocity, distance traveled, acceleration, and time required for objects in motion. Suitable for students in Grades 10-12.

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