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GERAK LURUS
Created by Fini husnul
JARAK = Total Lintasan yang ditempuh
Contoh Soal :
1. Seorang siswa berjalan 4 meter ke timur selama 1
sekon lalu berbelok ke utara sejauh 3 meter selama
1 sekon. Tentukan jarak dan perpindahan siswa
tersebut
2
3
meter
4 meter
Jarak = 4 meter + 3 meter = 7 meter
Perpindahan (yang sisi miring)
= 42 + 32
= 16 + 9
= 25
= 5 meter
PERPINDAHAN = jarak terpendek dari titik
awal ke titik akhir
laju dan kecepatan
Laju
Laju sesaat
Laju rata-rata
Kecepatan
3
4
× Laju adalah besarnya jarak yang ditempuh tiap satuan waktu. Laju termasuk
besaran skalar. Besarnya laju:
Laju =
𝐽𝑎𝑟𝑎𝑘
𝑤𝑎𝑘𝑡𝑢
× Laju sesaat adalah Laju gerak dalam rentang waktu yang sangat kecil.
Secara matematis, besarnya:
LajuSesaat =
𝑑𝑥
𝑑𝑡
1. keLAJUAN rata-rata
Kelajuan =
Jarak
Selang Waktu
5
V =
s
∆𝑡
2. Kecepatan rata-rata
6
Kecepatan rata-rata =
Perpindahan
Selang Waktu
𝑣 =
∆𝑥
∆𝑡
2. Sebuah benda bergerak dari A KE C melalui B. Panjang AB=50m, BC=30m, dan
waktu yang dibutuhkan dari A ke C adalah 20s. Hitunglah Kelajucan rata-rata dan
kecepatan rata-rata benda ?
7
Berdasarkan gambar diatas, kita peroleh jarak = 50 + 30 = 80 m
Dan perpindahan AC = 50 – 30 = 20m
Maka kelajuannya,
Kelajuan rata-rata =
𝐽𝑎𝑟𝑎𝑘
𝑤𝑎𝑘𝑡𝑢
8
=
80 𝑚
20 𝑠
= 40 m/s
Kecepatan Rata-rata =
∆𝑥
∆𝑡
=
20 𝑚
20 𝑠
= 1 𝑚/𝑠
Big
PercepatanRata-rata
9
10
𝒶 =
∆𝑣
∆𝑡
=
𝑣2−𝑣1
𝑡2−𝑡1
Keterangan:
a = percepatan rata-rata (m/s2)
Δv = perubahan kecepatan, yakni selisih kecepatan akhir dengan kecepatan
awal (V2 - V1).
Δt = selang waktu, yakni selisih waktu akhir dengan waktu awal (t2 - t1).
3. Seseorang mengendarai sepeda motor bergerak dengan kecepatan awal
15 m/s. Orang tersebut mempercepat laju kendaraannya sehingga dalam
waktu 10 sekon kecepatannya menjadi 20 m/s. Berapa percepatan sepeda
motor tersebut dalam satuan meter/sekon²?
11
Diketahui:
Kecepatan awal = 15 m/s.
Kecepatan akhir = 20 m/s
Maka, percepatannya, adalah 0,5 m/s
12
y = 2𝑥3
+ 12𝑥 + 3
𝒅𝒚
𝒅𝒙
= 2.3 𝑥3−1
+ 12
𝒅𝒚
𝒅𝒙
= 6 𝑥2
+ 12
Konsep matematika ; turunAN
13
Jika, y = 𝓪𝒙𝒏
Maka, turunan y terhadap x dapat
dituliskan ;
Contoh ;
𝒚′ =
𝒅𝒚
𝒅𝒙
= 𝓪𝒏. 𝑿𝒏−𝟏
y = 4 𝒙𝟑 r = 4𝑡2
+ 5t − 8
𝒅𝒓
𝒅𝒕
= 4.2 𝑡2−1
+ 5
𝒅𝒓
𝒅𝒕
= 8𝑡 + 5
𝒅𝒚
𝒅𝒙
= 4.3 x 3-1
𝒅𝒚
𝒅𝒙
= 12 𝑿𝟐
14
Yaitu kecepatan dan percepatan pada saat t (waktu) tertentu.
Posisi (r)
Kecepatan (v)
Percepatan (𝒶)
Turunan pertama posisi (r) terhadap waktu v =
𝑑𝑟
𝑑𝑡
Percepatan merupakan turunan pertama kecepatan terhadap waktu
atau turunan kedua posisi terhadap waktu.
15
Diketahui suatu benda bergerak dengan persamaan dibawah ini, tentukan
besar perccepatan benda pada detik ke-2
𝑥 = 4𝑡3
+ 2𝑡2
+ 3
𝑣(𝑡) = 4.3𝑡3−1
+ 2.2𝑡2−1
𝑣(𝑡) = 12𝑡2
+ 4t
𝑥 = 4𝑡3
+ 2𝑡2
+ 3
Jawab; percepatan merupakan turunan pertama dari keceptan, maka kita cari
dulu rumus kecepatannya.
Maka, turunan percepatannya;
𝑣(𝑡) = 12𝑡2
+ 4𝑡
𝒶 𝑡 = 12.2𝑡2−1
+ 4
𝒶 𝑡 = 24 𝑡 + 4
𝒶 2 = 24 (2) + 4
𝒶 𝑡 = 52 𝑚
𝑠2
Percepatannya 52 𝑚
𝑠2
Kerjakan
soal
berikut ini
16
NEXT
1. Sebuah mobil bergerak lurus ke timur sejauh 100 meter selama 4 sekon
lalu bergerak lurus ke barat sejauh 50 meter selama 1 sekon.
Tentukan kelajuan rata-rata dan kecepatan rata-rata mobil.
2. Perhatikan gambar di bawah! Seseorang mengelilingi tanah lapang
berbentuk persegi panjang. Panjang = 100 meter, lebar = 50 meter.
Setelah mengelilingi tanah lapang sebanyak tiga kali, orang tersebut
kembali ke posisi semula. Jika selang waktu tempuh = 100 sekon,
tentukan kelajuan rata-rata dan besar kecepatan rata-rata.
17
100 meter
50
meter
3. Seorang siswa berjalan 8 meter ke timur selama 2 sekon lalu berbelok ke
utara sejauh 6 meter selama 2 sekon. Tentukan kelajuan rata-rata dan besar
kecepatan rata-rata.
4. Diketahui sebuah mobil melaju dengan kecepatan awal yaitu 2 m/s. Setelah
mobil melaju 10 sekon, kecepatan mobil tersebut bertambah menjadi 4 m/s.
Berapa percepatan yang dimiliki oleh mobil tersebut ?
5. Diketahui suatu benda bergerak dengan persamaan dibawah ini, tentukan
besar posisi awal, kecepatan awal, dan percepatan benda tersebut.
𝒙 = 𝟓𝒕𝟐 + 𝟐𝟎𝒕 + 𝟓
18
THANKS!
Any questions?
You can find me at @googleclassroom
19
Created by Fini husnul

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Gerak lurus

  • 2. JARAK = Total Lintasan yang ditempuh Contoh Soal : 1. Seorang siswa berjalan 4 meter ke timur selama 1 sekon lalu berbelok ke utara sejauh 3 meter selama 1 sekon. Tentukan jarak dan perpindahan siswa tersebut 2 3 meter 4 meter Jarak = 4 meter + 3 meter = 7 meter Perpindahan (yang sisi miring) = 42 + 32 = 16 + 9 = 25 = 5 meter PERPINDAHAN = jarak terpendek dari titik awal ke titik akhir
  • 3. laju dan kecepatan Laju Laju sesaat Laju rata-rata Kecepatan 3
  • 4. 4 × Laju adalah besarnya jarak yang ditempuh tiap satuan waktu. Laju termasuk besaran skalar. Besarnya laju: Laju = 𝐽𝑎𝑟𝑎𝑘 𝑤𝑎𝑘𝑡𝑢 × Laju sesaat adalah Laju gerak dalam rentang waktu yang sangat kecil. Secara matematis, besarnya: LajuSesaat = 𝑑𝑥 𝑑𝑡
  • 5. 1. keLAJUAN rata-rata Kelajuan = Jarak Selang Waktu 5 V = s ∆𝑡
  • 6. 2. Kecepatan rata-rata 6 Kecepatan rata-rata = Perpindahan Selang Waktu 𝑣 = ∆𝑥 ∆𝑡
  • 7. 2. Sebuah benda bergerak dari A KE C melalui B. Panjang AB=50m, BC=30m, dan waktu yang dibutuhkan dari A ke C adalah 20s. Hitunglah Kelajucan rata-rata dan kecepatan rata-rata benda ? 7 Berdasarkan gambar diatas, kita peroleh jarak = 50 + 30 = 80 m Dan perpindahan AC = 50 – 30 = 20m Maka kelajuannya,
  • 8. Kelajuan rata-rata = 𝐽𝑎𝑟𝑎𝑘 𝑤𝑎𝑘𝑡𝑢 8 = 80 𝑚 20 𝑠 = 40 m/s Kecepatan Rata-rata = ∆𝑥 ∆𝑡 = 20 𝑚 20 𝑠 = 1 𝑚/𝑠
  • 10. 10 𝒶 = ∆𝑣 ∆𝑡 = 𝑣2−𝑣1 𝑡2−𝑡1 Keterangan: a = percepatan rata-rata (m/s2) Δv = perubahan kecepatan, yakni selisih kecepatan akhir dengan kecepatan awal (V2 - V1). Δt = selang waktu, yakni selisih waktu akhir dengan waktu awal (t2 - t1).
  • 11. 3. Seseorang mengendarai sepeda motor bergerak dengan kecepatan awal 15 m/s. Orang tersebut mempercepat laju kendaraannya sehingga dalam waktu 10 sekon kecepatannya menjadi 20 m/s. Berapa percepatan sepeda motor tersebut dalam satuan meter/sekon²? 11 Diketahui: Kecepatan awal = 15 m/s. Kecepatan akhir = 20 m/s Maka, percepatannya, adalah 0,5 m/s
  • 12. 12
  • 13. y = 2𝑥3 + 12𝑥 + 3 𝒅𝒚 𝒅𝒙 = 2.3 𝑥3−1 + 12 𝒅𝒚 𝒅𝒙 = 6 𝑥2 + 12 Konsep matematika ; turunAN 13 Jika, y = 𝓪𝒙𝒏 Maka, turunan y terhadap x dapat dituliskan ; Contoh ; 𝒚′ = 𝒅𝒚 𝒅𝒙 = 𝓪𝒏. 𝑿𝒏−𝟏 y = 4 𝒙𝟑 r = 4𝑡2 + 5t − 8 𝒅𝒓 𝒅𝒕 = 4.2 𝑡2−1 + 5 𝒅𝒓 𝒅𝒕 = 8𝑡 + 5 𝒅𝒚 𝒅𝒙 = 4.3 x 3-1 𝒅𝒚 𝒅𝒙 = 12 𝑿𝟐
  • 14. 14 Yaitu kecepatan dan percepatan pada saat t (waktu) tertentu. Posisi (r) Kecepatan (v) Percepatan (𝒶) Turunan pertama posisi (r) terhadap waktu v = 𝑑𝑟 𝑑𝑡 Percepatan merupakan turunan pertama kecepatan terhadap waktu atau turunan kedua posisi terhadap waktu.
  • 15. 15 Diketahui suatu benda bergerak dengan persamaan dibawah ini, tentukan besar perccepatan benda pada detik ke-2 𝑥 = 4𝑡3 + 2𝑡2 + 3 𝑣(𝑡) = 4.3𝑡3−1 + 2.2𝑡2−1 𝑣(𝑡) = 12𝑡2 + 4t 𝑥 = 4𝑡3 + 2𝑡2 + 3 Jawab; percepatan merupakan turunan pertama dari keceptan, maka kita cari dulu rumus kecepatannya. Maka, turunan percepatannya; 𝑣(𝑡) = 12𝑡2 + 4𝑡 𝒶 𝑡 = 12.2𝑡2−1 + 4 𝒶 𝑡 = 24 𝑡 + 4 𝒶 2 = 24 (2) + 4 𝒶 𝑡 = 52 𝑚 𝑠2 Percepatannya 52 𝑚 𝑠2
  • 17. 1. Sebuah mobil bergerak lurus ke timur sejauh 100 meter selama 4 sekon lalu bergerak lurus ke barat sejauh 50 meter selama 1 sekon. Tentukan kelajuan rata-rata dan kecepatan rata-rata mobil. 2. Perhatikan gambar di bawah! Seseorang mengelilingi tanah lapang berbentuk persegi panjang. Panjang = 100 meter, lebar = 50 meter. Setelah mengelilingi tanah lapang sebanyak tiga kali, orang tersebut kembali ke posisi semula. Jika selang waktu tempuh = 100 sekon, tentukan kelajuan rata-rata dan besar kecepatan rata-rata. 17 100 meter 50 meter
  • 18. 3. Seorang siswa berjalan 8 meter ke timur selama 2 sekon lalu berbelok ke utara sejauh 6 meter selama 2 sekon. Tentukan kelajuan rata-rata dan besar kecepatan rata-rata. 4. Diketahui sebuah mobil melaju dengan kecepatan awal yaitu 2 m/s. Setelah mobil melaju 10 sekon, kecepatan mobil tersebut bertambah menjadi 4 m/s. Berapa percepatan yang dimiliki oleh mobil tersebut ? 5. Diketahui suatu benda bergerak dengan persamaan dibawah ini, tentukan besar posisi awal, kecepatan awal, dan percepatan benda tersebut. 𝒙 = 𝟓𝒕𝟐 + 𝟐𝟎𝒕 + 𝟓 18
  • 19. THANKS! Any questions? You can find me at @googleclassroom 19 Created by Fini husnul