Question

A car starts from rest and acceleration uniformly for 10 s to a velocity of 8
m/s. It then run at a constant velocity and is finally brought to rest in 16 sec
with a constant retardation. The total distance covered by the car is 584m.
The value of retardation is​

Answer 1

Answer

Retardation = 0.5 m/s²

Given

A car starts from rest and acceleration uniformly for 10 s to a velocity of 8 m/s. It then run at a constant velocity and is finally brought to rest in 16 sec with a constant retardation. The total distance covered by the car is 584 m .

To Find

Value of retardation

Solution

Case - 1 : AB

Initial velocity , u = 0 m/s

[ ∵ Starts from rest ]

Time , t = 10 s

Final velocity , v = 8 m/s

Apply 1st equation of motion ,

⇒ v = u + at

⇒ (8) = (0) + a(10)

⇒ 8 = 0 + 10a

⇒ 10a = 8

⇒ a = 0.8 m/s²

Apply 3rd equation of motion ,

⇒ v² - u² = 2as

⇒ (8)² - (0)² = 2(0.8)s

⇒ 64 - 0 = 1.6s

⇒ 64 = 1.6s

⇒ 1.6s = 64

⇒ s (AB) = 40 m

Distance , BC = 584 - 40

⇒ Distance , AB = 544 m

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Case - 2 : BC

Distance , s = 544 m

Time , t = 16 s

Initial velocity , u = 8 m/s

Final velocity , v = 0 m/s

Retardation , a = ? m/s²

Apply 1 st equation of motion ,

⇒ v = u + at

⇒ (0) = (8) + a(16)

⇒ 0 = 8 + 16a

⇒ 16a = -8

⇒ a = - 0.5 m/s²

So , retardation , a = - 0.5 m/s²

Note : - ve sign denotes retardation

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