Question

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Answer 1

Given , mass of the car (m) = 1000 kg

(i) As, shown in graph,

Distance covered in first 2 sec

= Area of Δ inscribed in first 2 s.

= 1 / 2 × base × height

= 1 / 2 × 2 ×15 = 15 m

(ii) Force (F) = mass (m) × acceleration (a)

Time taken by force to stop car = 6 - 5 = 1

second

Initial velocity (i.e. velocity at B) = 15 m/s

Final velocity (at point C) = 0 m/s

using equation v = u + at


a = (v-u) / t = (0 -20) / 0.02 = -15 m/s2

(-ve sign indicates that it is retardation )

Force applied by brakes = 1000 × -15 = -15000 N

= - 15 KN

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Answer 2

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Given ,

mass of the car (m) = 1000 kg

(i) As, shown in graph,

Distance covered in first 2 sec

= Area of Δ inscribed in first 2 s.

= 1 / 2 × base × height

= 1 / 2 × 2 ×15 = 15 m

(ii) Force (F) = mass (m) × acceleration (a)

Time taken by force to stop car = 6 - 5 = 1

second

Initial velocity (i.e. velocity at B) = 15 m/s

Final velocity (at point C) = 0 m/s

using equation v = u + at

a = (v-u) / t = (0 -20) / 0.02 = -15 m/s2

(-ve sign indicates that it is retardation )

Force applied by brakes = 1000 × -15 = -15000 N

= - 15 KN