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

Explanation:

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Answer 2

Given:

The acceleration, a1 = 5 m/s²

The deceleration, a2 = -20 m/s²

Total distance, (s1 + s2) = 200 m

To Find:

The maximum velocity attained by the scooter.

Calculation:

- For acceleration, we have u = 0

- Applying 3rd equation of motion, we get:

v² = u² + 2a1s1

⇒ v² = 0 + 2 × 5 × s1

⇒ v² = 10 s1  ........(i)

- For deceleration, we have:

V = 0 and U = v

- Applying 3rd equation of motion, we get:

V² = U² + 2a2s2

⇒ 0 = v² + 2 × (-20) × s2

⇒ v² = 40 s2  

Putting the value of (i), we get:

10 s1  = 40 s2

⇒ 10 s1 - 40 s2 = 0 ........(ii)

- Also, we have (s1 + s2) = 200 ......(iii)

- On solving (ii) and (iii), we get:

s1 = 160 m and s2 = 40 m

- Putting it in (i), we get:

v² = 10 × 160

⇒ v = √1600

⇒ v = 40 m/s

- So, the maximum velocity attained by the scooter is​ 40 m/s.