4, 5, 6, and 7 Numbered Physics Questions Explanation and Solutions

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Kübra, sorularını tek tek adım adım çözelim.

4. İvme–zaman ve hız–zaman grafikleri

Verilenler:

  • Başlangıç hızı: v_0 = 10\ \mathrm{m/s}
  • İlk 10 s süreyle ivme: a_1 = +3\ \mathrm{m/s^2}
  • Sonraki 5 s süreyle ivme: a_2 = -2\ \mathrm{m/s^2}

a) İvme–zaman grafiği:

  • 0 \le t < 10\ \mathrm{s} aralığında a = +3\ \mathrm{m/s^2} (sabit yatay çizgi).
  • 10 \le t \le 15\ \mathrm{s} aralığında a = -2\ \mathrm{m/s^2} (sabit yatay çizgi).

b) Hız–zaman grafiği:

  • Başlangıçta v(0)=10\ \mathrm{m/s}.
  • İlk 10 s’de: v = v_0 + a_1\,t = 10 + 3\,t. Nokta olarak $t=10$’da v(10)=10+30=40\ \mathrm{m/s}.
  • Son 5 s’de: v = 40 + (-2)\,(t-10) = 40 -2\,(t-10). $t=15$’te v(15)=40-2\cdot5=30\ \mathrm{m/s}.

Yani grafik:

  • $0\to10,$s arası pozitif eğimle artan doğru (10’dan 40’a).
  • $10\to15,$s arası negatif eğimle azalan doğru (40’dan 30’a).

5. Ay’da tüy ve çekicin serbest düşüşü

Ay’da hava direnci ihmal edildiğine göre tüm cisimler aynı sabit ivme (g_{Ay}) ile düşer.

a) İvmelerin karşılaştırılması

  • Tüyün ivmesi = Çekicin ivmesi = g_{Ay} (eşit ve cisim bağımsız).

b) Tüyün A, B, C noktalarındaki hız sıralaması

  • Serbest düşmede hız zamanla artar. Dolayısıyla V_A < V_B < V_C.

c) Tüy ve çekicin C noktasındaki hızları

  • Aynı zamanda ve aynı ivme ile düştükleri için v_{\text{tüy},C} = v_{\text{çekici},C}.

6. 180 m yükseklikten serbest düşme

Verilen: h=180\ \mathrm{m}, v_0=0, g=10\ \mathrm{m/s^2}.

a) Zamanı bulmak için
h = \tfrac12\,g\,t^2 \quad\Rightarrow\quad t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2\cdot180}{10}} = \sqrt{36} = 6\ \mathrm{s}.

b) Çarpma hızı
v = v_0 + g\,t = 0 + 10\cdot6 = 60\ \mathrm{m/s}
(şağı doğru).

7. Yatay atışta yatay hızın değişimi

Hava direnci yoksa yatay eksende net kuvvet yoktur. Newton’un birinci yasasına göre yatay doğrultudaki hız bileşeni sabit kalır. Dolayısıyla:

• Yatay hız değişmez, çünkü o doğrultuda bir kuvvet (direnç) etkisi yoktur.

Özet:

  1. Soru 4:

    • a–t: 0\!-\!10\,\mathrm{s} arası +3, 10\!-\!15\,\mathrm{s} arası –2.
    • v–t: 0’dan 10 s’ye 10→40 m/s, sonra 40→30 m/s.

  2. Soru 5:

    • (a) İvmeler eşittir.
    • (b) V_A < V_B < V_C.
    • (c) v_{\text{tüy},C} = v_{\text{çekici},C}.

  3. Soru 6:

    • Zaman: 6 s.
    • Çarpma hızı: 60 m/s.

  4. Soru 7:

    • Yatay hız sabit kalır (yatayda net kuvvet yok).
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4, 5, 6, and 7 Numbered Physics Questions Explanation and Solutions

Let’s analyze and solve the physics problems step by step.

4. Positive Direction Motion with Given Velocity and Acceleration

Problem:

  • Initial velocity v_0 = 10 \, m/s (positive direction)
  • At t=0, acceleration a = 3 \, m/s^2
  • Acceleration changes to 2 \, m/s^2 after some time
  • The object moves with this acceleration for 5 seconds
  • Draw velocity-time and acceleration-time graphs for the motion between t=0 and t=10 seconds

Solution:

  • From t=0 to t=5 s, acceleration a = 3 \, m/s^2
  • From t=5 to t=10 s, acceleration a = 2 \, m/s^2

Velocity-time graph:

  • Velocity at t=0 is 10 \, m/s
  • Velocity increases linearly with slope = acceleration
  • For 0 \leq t \leq 5:
    v(t) = v_0 + a t = 10 + 3t
    At t=5,
    v(5) = 10 + 3 \times 5 = 25 \, m/s
  • For 5 < t \leq 10:
    v(t) = v(5) + 2 (t-5) = 25 + 2(t-5)
    At t=10,
    v(10) = 25 + 2 \times 5 = 35 \, m/s

Acceleration-time graph:

  • Constant at 3 \, m/s^2 from 0 to 5 seconds
  • Then constant at 2 \, m/s^2 from 5 to 10 seconds

5. Free Fall on the Moon’s Atmosphere

Problem:

  • Moon’s atmosphere is negligible (no air resistance)
  • Three objects (feather, arrow, and a hammer) are dropped from the same height simultaneously
  • Tasks:
    a) Compare accelerations of feather and arrow
    b) Compare velocities of feather, arrow, and hammer
    c) Compare velocities of feather and arrow in vacuum

Solution:

a) Acceleration comparison:

  • Since air resistance is negligible, all objects fall with the same acceleration due to gravity on the Moon.
  • Therefore,
    a_{\text{feather}} = a_{\text{arrow}} = a_{\text{hammer}} = g_{\text{moon}}

b) Velocity comparison:

  • Since all objects fall freely without air resistance, their velocities at any time are equal.
  • So,
    v_{\text{feather}} = v_{\text{arrow}} = v_{\text{hammer}}

c) Velocities in vacuum:

  • In vacuum, no air resistance exists.
  • Feather and arrow fall with the same acceleration and velocity at any time.
  • So,
    v_{\text{feather}} = v_{\text{arrow}}

6. Ball Drop from 180 m Height (Ignoring Air Resistance)

Problem:

  • Orhan drops a ball from 180 m height (initial velocity zero)
  • Gravity g = 10 \, m/s^2
  • Find:
    a) Time to hit the ground
    b) Velocity just before impact

Solution:

a) Time to hit the ground:

Using the equation of motion for free fall:

s = \frac{1}{2} g t^2

Given s = 180 \, m, solve for t:

180 = \frac{1}{2} \times 10 \times t^2 \implies 180 = 5 t^2 \implies t^2 = \frac{180}{5} = 36 \implies t = 6 \, s

b) Velocity before impact:

v = g t = 10 \times 6 = 60 \, m/s

7. Horizontal Velocity Change in Projectile Motion with Air Resistance

Problem:

  • In an environment with air resistance, a projectile is thrown horizontally from a height
  • Explain why the horizontal velocity does not change until it hits the ground

Explanation:

  • In reality, air resistance affects horizontal velocity, causing it to decrease.
  • However, if air resistance is negligible or ignored, no horizontal force acts on the projectile.
  • According to Newton’s first law, velocity remains constant if no force acts.
  • Therefore, horizontal velocity remains constant during the fall.
  • The vertical velocity changes due to gravity, but horizontal velocity remains unchanged.

Summary Table

Question Key Points Formula/Result
4 Velocity changes with piecewise acceleration v(t) = 10 + 3t (0-5s), v(t) = 25 + 2(t-5) (5-10s)
5a All objects have same acceleration on Moon a_{\text{feather}} = a_{\text{arrow}} = g_{\text{moon}}
5b Velocities equal due to no air resistance v_{\text{feather}} = v_{\text{arrow}} = v_{\text{hammer}}
5c Velocities equal in vacuum v_{\text{feather}} = v_{\text{arrow}}
6a Time to hit ground from 180 m t = 6 \, s
6b Velocity before impact v = 60 \, m/s
7 Horizontal velocity constant without air resistance No horizontal force → constant velocity

If you want, I can also help you draw the velocity-time and acceleration-time graphs for question 4 or explain any part in more detail.

@Kubr