Question : The relation between time t and distance x is t=ax2+bx where a and b are constants. Express the instantaneous acceleration in terms of instantaneous velocity.
Doubt by Riddhi
Solution :
Note : Advance concepts of differentiation like chain rule is used to solve this question.
t=ax2+bx — ⭐
t=ax2+bx — ⭐
differentiating both sides w.r.t t
dt/dt = (2ax)×(dx/dt)+b(dx/dt)
1 = (2ax)(v)+b(v)
1 = (2ax)(v)+b(v)
1=v[2ax+b]
1/[2ax+b] = v
v = [2ax+b]-1 — (1)
Again differentiating both sides w.r.t. t
dv/dt = (-1)[2ax+b]-2×2a[dx/dt]
A = -2a[(2ax+b)-1]2×v
A=-2a[v]2×v
A=-2av3
A=-2a[v]2×v
A=-2av3
Note : You can make it little bit more simpler if you differentiate the equation ⭐ both sides w.r.t. to x instead of t.