Question

A car of mass 200kg moving at 36 km /h is brought to rest after it covered a distance of 10 m.find the retarding force action on the car​

Answer 1

Answer:

Answer:

Mass of the car (m) = 200 kg Initial speed (u) = 36 km/h = 10 m/s Final velocity (v) = 0

Distance covered (S) = 10 m

v2 - u2 = 2aS 0 - 100 = 2 x a x 10 - 100 = 20 aa = - 100/20 = -5 m/s2

F = ma = 200 x 5 = -1000 N

Retarding force = 1000 N

Answer 2

QUESTION

A car of mass 200kg moving at 36 km /h is brought to rest after it covered a distance of 10 m.find the retarding force action on the car.

✴SOLUTION

❇GIVEN

• Mass of car = 200kg.
• Initial velocity of car = 36 km/h = 10m/s
• Final velocity of car = 0 ( as it comes to rest. )
• Distance = 10m

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✳EQUATION OF MOTION

$1. v = u + at$

$2. S = ut + \frac{1}{2}at²$

$3. v² = u² - 2aS$

Where,

• v = final velocity
• u = initial velocity
• t = time
• S = distance
• a = acceleration

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Using 3rd equation of motion.

$v² = u² - 2aS$

$0² = 10² - (2×a×10)$

$0 = 100 - 20a$

$20a = 100$

$a = \frac{100}{20}$

$a = 5m/s²$

Therefore, acceleration = 5m/s²

Retardation = -5m/s²

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NOW

$Retarding \: force = 200 × -5$

$F = -1000N$

$F = -10³N$

RETARDING FORCE = -10³N