The relation between the’t’ and distance 'x' is t = ax2+bx, where a & b are constants. The retardation is: (a) 2a.v3 (b) 2b.v3 (c) 2ab.v3 (d) 2b2v3

Question

The relation between the’t’ and distance 'x' is

$t = ax^2 + bx,$
where a & b are constants. The retardation is:

(a) 2a.v3
(b) 2b.v3
(c) 2ab.v3
(d) 2b²v3

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Solution

The given relation is t = ax^2 + bx.

First, we need to find the velocity 'v'. Velocity is the rate of change of distance with respect to time. So, we differentiate 'x' with respect to 't' to get velocity.

dx/dt = v = 2ax + b

Next, we need to find the retardation or deceleration, which is the rate of change of velocity with respect to time. So, we differentiate 'v' with respect to 't' to get retardation.

dv/dt = a(2) = 2a

So, the retardation is 2a, which is not listed in the options. Therefore, there might be a mistake in the question or the options provided.

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The relation between the’t’ and distance 'x' is t = ax2+bx, where a & b are constants. The retardation is: (a) 2a.v3 (b) 2b.v3 (c) 2ab.v3 (d) 2b2v3

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