Question

The relation between time y and distance x is t = ax² + bx , where a and b are constants . the acceleration is [AIEEE 2005]
(a) -2abv² (b) 2bv³ (c) -2av³ (d) 2av² ​

Answer 1

Explanation:- Given t = ax² +bx

Differentiating it w.r.t t , we get

• dt/dx = 2ax dx/dt + b dx/dt
• V = dx/dt = 1/(2ax + b) _____(1)
• Again , differentiating w.r.t to t we get
• d²x/dt² = -2a/(2ax+b)² dx/dt
• ∴ a = d²x/dt²
• = -1/(2ax+b)²•2a/(2ax+b)
• or , a = -2a/(2ax+b)²
• => a = -2av³ ___(from 1)

Since , acceleration is -2av³.

Option (c) is correct option

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Answer 2

Answer:

C

Explanation:

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