Question

17. A car accelerates uniformly from 18 kmh to 36 kmh in 2 seconds.
Calculate.
(a) Acceleration
(b) Distance covered by the car in that time.
hnunht to rest after covering a di​

Answer 1

Answer:

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Answer 2

GivEn:

• Initial velocity of Car, u = 18 km/h = 18 × 5/18  m/s = 5 m/s
• Final velocity of Car, v = 36 km/h = 36 × 5/18 m/s = 10 m/s
• Time taken for changing velocity, t = 2 sec

To find:

• Acceleration of Car, a =?
• Distance covered by Car in that time, s =?

Formula required:

• First equation of motion

     v = u + a t

• Second equation of motion

     s = u t + 1/2 a t²

[ Where v is final velocity, u is initial velocity, a is acceleration, t is time taken, s is distance covered ]

Solution:

Using first equation of motion

→ v = u + a t

→ ( 10 ) = ( 5 ) + a ( 2 )

→ 10 - 5 = 2 a

→ 5 = 2 a

→ a = 5 / 2

→ a = 2.5 m/s²

Therefore,

• Acceleration of Car is 2.5 m/s².

Using Second equation of motion

→ s = u t + 1/2 a t²

→ s = ( 5 ) ( 2 ) + 1/2 ( 2.5 ) ( 2 )²

→ s = 10 + 5

→ s = 15 m

Therefore,

• Distance covered by the Car in that time is 15 m.