Question

a. The driver of a moving car suddenly sees ahead a very old man on the middle of the road. To save the old man he applies the brake, which produces an acceleration of 4 m/s^2 in opposite direction to the motion. After three seconds the car comes to rest. (i) Calculate the initial velocity and the distance traveled by the car before stopping. (ii) If the old man is 20 m away from the car, would the driver succeed to save the old man? Justify b. A body is thrown vertically upward with a velocity 'u'. What will be the greatest height 'h' the body can attain?​

Answer 1

Explanation:

If question b has only that much of information then you can directly assign it to the eqn.

s=ut-1/2gt^2

Answer 2

Solution (a)

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Given :-

• final velocity of Car, v = 0

(since, Car comes to rest)

• acceleration of Car , a = - 4 ms⁻²

(negative acceleration because acceleration produced is opposite to the direction of motion)

• time taken by Car to stop , t = 3 s

To find :-

• initial velocity of Car , u = ?
• distance traveled by Car after applying brakes and before stopping , s = ?
• If the old man is 20 m away from the Car, would driver succeed to save the old man.

Answer & Explanation :-

Using first equation of motion

→ v = u + at

putting known values

⇢0 = u + (-4) (3)

⇢0 = u - 12

⇢ u = 12 ms⁻¹

Hence, initial velocity of Car is 12 ms⁻¹ .

Using second equation of motion

→ s = ut + (1/2) at²

putting known values

⇢ s = (12)(3) + (1/2) (-4)(3)²

⇢ s = 36 + (-18)

⇢ s = 36 - 18

⇢ s = 18 m

Hence, Distance traveled by Car is 18 m before stopping and after applying brakes.

Now,

Since , Man is 20 m away from the Car and Car traveled 18 m to come to rest it means Driver would succeed in saving Old man By stopping his Car 2 metres away from the man.

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Solution (b)

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Given :-

• A body is thrown vertically upward with a velocity u

To find :-

• Greatest height 'h' the body could attain

Answer & Explanation :-

↦initial velocity of body = u (given)

↦final velocity of body = v = 0

(when the object will reach maximum height h then final velocity become zero)

↦acceleration of body = a = - g  

(acceleration due to gravity and body is moving upward)

↦distance covered by body  = h

Using third equation of motion

→ 2 a h = v² - u²

Putting known values

⇢ 2 (-g) h = (0) - u²

⇢ - 2 g h = - u²

⇢ 2 g h = u²

⇢ h = u² / 2 g

Hence, the greatest height attained by body will be u² / 2 g  where u is the initial velocity of body and g is the acceleration due to gravity.

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