مراجعة رياضيات تطبيقية لغات للثانوية العامة 2026.. راجع قبل الامتحان

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مراجعة رياضيات تطبيقية لغات 3 ثانوي 2026

9] A car's engine does work at a constant rate of magnitude 5 kilowatt and the car's mass is 1200 kg. If the car moves on a horizontal road against a constant resistance of magnitude 325 newtons, find:

• a] The magnitude of the car's acceleration when its velocity is 8m/sec.​​​​​​​

• b] The car's maximum velocity.

The solution

a]

$$\text{The power} = F \times v$$

$$\therefore 5 \times 1000 = F \times 8$$

$$\therefore F = 625 \text{ Newton} \quad \therefore F - R = ma$$

$$\therefore 625 - 325 = 1200 \times a$$

$$\therefore a = \frac{1}{4} \text{ m/sec}^2$$

b] At maximum velocity:

$$F = R = 325$$

$$\therefore 5 \times 1000 = 325 \times v$$

$$\therefore v = \frac{200}{13} \text{ m/sec}$$

تفاصيل الملف الثاني: 748419134_2549210405528664_7973398858853791868_n.jpg

and the two bodies are in one horizontal plane?

The solution

$$\because 125 \times 980 - T = 125 a \quad \text{-------- (1)}$$

$$T - 120 \times 980 = 120 a \quad \text{-------- (2)}$$

Adding (1) and (2)

$$\therefore a = \frac{5 \times 980}{245} = 20 \text{ cm/sec}^2$$

from (2) :

$$\therefore T = 120 \times 980 + 120 \times 20 = 120000 \text{ dyne}$$

$$\therefore p = 2T = 240000 \text{ dyne}$$

$$v_0 = 0 \quad \& \quad t = 1 \quad \& \quad a = 20 \text{ cm/sec}^2$$

$$S = v_0 t + \frac{1}{2} a t^2$$

$$= 0 + \frac{1}{2} \times 20 \times 1 = 10 \text{ cm}$$

$$\therefore \text{The vertical distance between the two bodies} = 20 \text{ cm.}$$