Question

8. A scooter acquires a velocity of 36km/hr in 10seconds just after the start. It takes 20

seconds to stop. Calculate the acceleration in the two cases.

Answer 1

may be this is your answer.
Hope it surely helped u

Answer 2

Greetings,
The answer to your question is typed below↓
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Given:
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Case 1: ( after start )
initial velocity [ u ]  = 0
final velocity [ v ] = 36 km/hr = 10 m/s  [ ∵ 1 km/hr = 5/18 m/s]
time [ t ] = 10 s
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Case 2: ( Before Stoppin')
u = 10m/s
v =0
t = 20 s 
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Sol: (For 'Case 1')

Substituting the values of 'Case 1' in the 1st equation of motion; v=u+at
We get↓

⇒10 = 0 + a x 10
⇒ a = 1 m/s²
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Sol: (For 'Case 2')

Similarly:

For Retarding bodies, v = u - a x t
∴ Substituting values for 'Case 2' in the equation;
We get : ↓

⇒ 0 = 10 - a x 20
⇒ a x 20 = 10
⇒ a = 10/20
⇒ a= 1/2
⇒ a = 0.5 m/s²

∴
Acceleration in the 1st case       = 1 m/s²
Retardation in the second case = 0.5 m/s²
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P.S: Enjoy ;)