Question

the relation between time t and distance x is t = ax square + bx where a and b are constants.find acceleration ?

Answer 1

Answer on Question #42100 – Physics - Mechanics | Kinematics | Dynamics

the relation between time $t$ and distance $x$ is $t = ax$ square + bx where $a$ and $b$ are constants. find acceleration?

Solution:

$$t = a x ^ {2} + b x$$

Differentiate both side with respect to time $t$ (velocity $v = \frac{dx}{dt}$):

$$\frac {d t}{d t} = 2 a x \frac {d x}{d t} + b \frac {d x}{d t}$$ $$1 = 2 a x \cdot v + b \cdot v$$ $$v = \frac {1}{2 a x + b} = (2 a x + b) ^ {- 1}$$

Again differentiate both side with respect to time $t$ (acceleration $a = \frac{dv}{dt}$)

$$\frac {d v}{d t} = - 2 a \frac {d x}{d t} (2 a x + b) ^ {- 2} = - \frac {2 a \cdot v}{(2 a x + b) ^ {2}} = - \frac {2 a}{(2 a x + b) ^ {3}} = a$$

Answer: acceleration is equal to $-\frac{2a}{(2ax + b)^3}$

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