Scbuah bendo mula-mula diam, kemudian bergerak dengan percepatan 2m/s2 solama 4 detik, solanjutnya bergerak dongan kocepstan konstan solama 20 detik selanjutnya direm dengan perlambatan 4m/s2 sampai berhenti a. Gambarkan gerakan benda tersebut dalam gravik v-(kecepatan-waktu) b. Jarak total yang ditempuh. c. Kecepatan rata-rata benda tersebut Select one: a.14 m b.84 m c. 184 m d.18 m

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Scbuah bendo mula-mula diam, kemudian bergerak dengan percepatan 2m/s2 solama 4 detik, solanjutnya bergerak dongan kocepstan konstan solama 20 detik selanjutnya direm dengan perlambatan 4m/s2 sampai berhenti a. Gambarkan gerakan benda tersebut dalam gravik v-(kecepatan-waktu) b. Jarak total yang ditempuh. c. Kecepatan rata-rata benda tersebut Select one: a.14 m b.84 m c. 184 m d.18 m

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Answer 1

Here's a step-by-step solution to the problem:

a. Velocity-Time Graph:

The motion of the object can be divided into three phases:

* Phase 1 (Acceleration): The object starts from rest (v=0 m/s) and accelerates at 2 m/s² for 4 seconds. The final velocity at the end of this phase can be calculated using the equation: v = u + at, where 'v' is final velocity, 'u' is initial velocity, 'a' is acceleration, and 't' is time. Therefore, v = 0 + (2 m/s²)(4 s) = 8 m/s.

* Phase 2 (Constant Velocity): The object moves at a constant velocity of 8 m/s for 20 seconds.

* Phase 3 (Deceleration): The object decelerates at 4 m/s² until it comes to a stop (v=0 m/s). The time it takes to stop can be calculated using the same equation as before: 0 = 8 m/s + (-4 m/s²)t. Solving for t, we get t = 2 seconds.

The velocity-time graph would look like this:

```
Velocity (m/s)
^
| /
| /
8|-----/-------->
| /
| /
| /
| /
|/
+------------------------> Time (s)
0 4 24 26

```

The graph shows a linearly increasing velocity from 0 to 8 m/s over 4 seconds, a horizontal line at 8 m/s for 20 seconds, and a linearly decreasing velocity from 8 m/s to 0 m/s over 2 seconds.

b. Total Distance:

The total distance is the sum of the distances covered in each phase. We can calculate the distance using the following equations:

* Phase 1: Distance = ut + (1/2)at² = 0 + (1/2)(2 m/s²)(4 s)² = 16 m

* Phase 2: Distance = velocity × time = (8 m/s)(20 s) = 160 m

* Phase 3: Distance = ut + (1/2)at² = (8 m/s)(2 s) + (1/2)(-4 m/s²)(2 s)² = 8 m

* Total Distance: 16 m + 160 m + 8 m = 184 m

c. Average Velocity:

Average velocity is the total displacement divided by the total time. In this case, the total displacement is the total distance (since the motion is all in one direction), and the total time is 4 s + 20 s + 2 s = 26 s.

Average velocity = Total distance / Total time = 184 m / 26 s ≈ 7.08 m/s

Therefore, the correct answer is c. 184 m for the total distance. The average velocity is approximately 7.08 m/s.

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