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:

40 m/s

Explanation:

Using the third equation of motion,

v² = u² + 2as

Here initial velocity is 0

Hence

v² = 0 + 2 × 5 × s1

v² = 10 s1  ........(a)

Incase of deceleration, we have

V = 0 and U = v

V² = U² + 2a2s2

0 = v² + 2 × (-20) × s2

v² = 40 s2  

Putting the value of (a), we get:

10 s1  = 40 s2

10 s1 - 40 s2 = 0 ........(b)

(s1 + s2) = 200 ......(c)

Equating b and c we get

s1 = 160 m and s2 = 40 m

v² = 10 × 160

v = 40 m/s

Answer 2

maximum velocity attained by the scooter = 40 m/s

Explanation:

Let say maximum velocity attained  = V

using V² - u² = 2aS

V² - 0 = 2(5)S₁

=> S₁  = V²/10

0² - V² = 2(-20)S₂

=> S₂ = V²/40

S₁  + S₂ = 200

=> V²/10 + V²/40  = 200

=> 4V² + V² = 200 * 40

=> 5V² = 200 * 40

=> V² = 40 * 40

=> V = 40

maximum velocity attained by the scooter = 40 m/s

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