Mechanics> Motion in a straight line...
The relation between time t and position x along x-axis is given by t = ax^2 + bx, where a and b are alpha and beta and they are constants. Prove that the retardation is 2aV^3, where V is the velocity at time t.
2 Answers
t = ax^2 + bx..............1
differentiate eq 1 wrt t
dt/dt = adx^2/dt + bdx/dt (dx/dt =V )
1=a(2xV) + bV...........2
again differnetiate wrt t
0=2a(xA+V^2) + bA (dV/dt =A (retardation) )
A=-2Av^2 /2ax+b.................3
put the value of x in 3
A=-2Av^3
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