Question

A scooter accelerates from rest at a constant rate of 5 m/s2 for some time then it decelerates at a constant rate of 20 m/s2 and comes to rest after covering the total distance of 200 m. The maximum velocity attained by the scooter is

Answer 1

Answer:

to find v

Explanation:

here a=5m/s*2. ,redardation=20m/s*2. v=u+at. v=15/200. 200/15. v=14m/s hope may be right I am not sure

Answer 2

using the third equation of motion

b square is equal to Usquare + 2 a s

initial velocity is zero

hence,

v square is equal to 0 + 2 into 5 into S1

v square is equal to 10 S1 .... (a)

in case of declaration we have

v is = zero and u = v

v square= v square+ 2a 2s 2

o = v square + 2*(-20)*S2

v square 40 S2

putting the value of (a) we get

10 rs 1 is equal to 40 S2

10 rs 1 - 4o s2 is equal to zero...b

(S1) + S2 =200...c

equating b and c we get

s1 is equal to 160m and S2 is equal to 40 metre

v square = 10*160

v=40m/s

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