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41. Calculating final velocity with multiple phases:

A car accelerates from 0 to 20 m/s in 4 seconds, then continues at this speed for 5 seconds, and finally decelerates to 0 in 3 seconds. What is the total distance traveled?

Sagot :

Let's break down the car's journey into three distinct phases and calculate the distance covered in each phase. Finally, we'll sum all the distances for the total distance travelled.

### Phase 1: Acceleration from 0 to 20 m/s in 4 seconds

1. Initial velocity (\( u_1 \)): 0 m/s
2. Final velocity (\( v_1 \)): 20 m/s
3. Time (\( t_1 \)): 4 seconds

To find the acceleration (\( a_1 \)) during this phase, we can use the equation:

[tex]\[ a_1 = \frac{v_1 - u_1}{t_1} = \frac{20 \, \text{m/s} - 0 \, \text{m/s}}{4 \, \text{s}} = 5 \, \text{m/s}^2 \][/tex]

Now, using the formula for distance (\( s_1 \)):

[tex]\[ s_1 = u_1 t_1 + \frac{1}{2}a_1 t_1^2 \][/tex]

Substituting in the known values:

[tex]\[ s_1 = 0 \times 4 + \frac{1}{2} \times 5 \times 4^2 \][/tex]
[tex]\[ s_1 = 0 + \frac{1}{2} \times 5 \times 16 \][/tex]
[tex]\[ s_1 = 0 + 40 \][/tex]
[tex]\[ s_1 = 40 \, \text{meters} \][/tex]

### Phase 2: Constant speed at 20 m/s for 5 seconds

1. Velocity (\( v_2 \)): 20 m/s
2. Time (\( t_2 \)): 5 seconds

Distance (\( s_2 \)) in this phase can be found using the formula:

[tex]\[ s_2 = v_2 t_2 \][/tex]

Substituting the values:

[tex]\[ s_2 = 20 \, \text{m/s} \times 5 \, \text{s} \][/tex]
[tex]\[ s_2 = 100 \, \text{meters} \][/tex]

### Phase 3: Deceleration from 20 m/s to 0 in 3 seconds

1. Initial velocity (\( u_3 \)): 20 m/s
2. Final velocity (\( v_3 \)): 0 m/s
3. Time (\( t_3 \)): 3 seconds

To find the deceleration (\( a_3 \)):

[tex]\[ a_3 = \frac{v_3 - u_3}{t_3} = \frac{0 \, \text{m/s} - 20 \, \text{m/s}}{3 \, \text{s}} = -\frac{20}{3} \, \text{m/s}^2 \approx -6.67 \, \text{m/s}^2 \][/tex]

Now, using the formula for the distance (\( s_3 \)):

[tex]\[ s_3 = u_3 t_3 + \frac{1}{2} a_3 t_3^2 \][/tex]

Substituting the values:

[tex]\[ s_3 = 20 \times 3 + \frac{1}{2} \times \left(-\frac{20}{3}\right) \times 3^2 \][/tex]
[tex]\[ s_3 = 60 + \frac{1}{2} \times -20 \times 3 \][/tex]
[tex]\[ s_3 = 60 - 30 \][/tex]
[tex]\[ s_3 = 30 \, \text{meters} \][/tex]

### Total Distance Travelled

The total distance travelled by the car is the sum of the distances covered in all three phases:

[tex]\[ \text{Total Distance} = s_1 + s_2 + s_3 \][/tex]

Substitute the values:

[tex]\[ \text{Total Distance} = 40 \, \text{m} + 100 \, \text{m} + 30 \, \text{m} \][/tex]
[tex]\[ \text{Total Distance} = 170 \, \text{meters} \][/tex]

Thus, the total distance travelled by the car is 170 meters.
I spent a while on this, not sure if it’s correct but here:

To determine the total distance traveled by the car, we'll break the motion into three phases: acceleration, constant speed, and deceleration.

Phase 1: Acceleration
- Initial velocity (u1): 0 m/s
- Final velocity (v1): 20 m/s
- Time (t1): 4 seconds

Acceleration (a1) can be calculated as:
a1 = (v1 - u1) / t1
a1 = (20 m/s - 0 m/s) / 4 s
a1 = 5 m/s²

Distance (d1) during this phase can be calculated using the formula:
d1 = u1 * t1 + 0.5 * a1 * t1²
d1 = 0 * 4 + 0.5 * 5 * 4²
d1 = 0 + 0.5 * 5 * 16
d1 = 40 meters

Phase 2: Constant speed
- Speed (v2): 20 m/s
- Time (t2): 5 seconds

Distance (d2) during this phase can be calculated as:
d2 = v2 * t2
d2 = 20 m/s * 5 s
d2 = 100 meters

Phase 3: Deceleration
- Initial velocity (u3): 20 m/s
- Final velocity (v3): 0 m/s
- Time (t3): 3 seconds

Deceleration (a3) can be calculated as:
a3 = (v3 - u3) / t3
a3 = (0 m/s - 20 m/s) / 3 s
a3 = -20/3 m/s²

Distance (d3) during this phase can be calculated using the formula:
d3 = u3 * t3 + 0.5 * a3 * t3²
d3 = 20 * 3 + 0.5 * (-20/3) * 3²
d3 = 60 - 0.5 * 20 * 3
d3 = 60 - 30
d3 = 30 meters

Total distance traveled is the sum of the distances for all three phases:
Total distance = d1 + d2 + d3
Total distance = 40 meters + 100 meters + 30 meters
Total distance = 170 meters

The total distance traveled by the car is 170 meters.
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