Question 90: Motion in 1D

Question

The relation between time t and distance x is t = ax2 + bx where a and b are constants.
The acceleration is

Accepted Answer

The relationship between time $ t $ and distance $ x $ is defined as $ t = ax^2 + bx $, where $ a $ and $ b $ are constants.
To find the acceleration, follow these steps:
First, differentiate the equation with respect to time $ t $:
$ \frac{d}{dt}(t) = a \frac{d}{dt}(x^2) + b \frac{dx}{dt} $
This simplifies to:
$ 1 = a \cdot 2x \frac{dx}{dt} + b \frac{dx}{dt} $
Given $ \frac{dx}{dt} = v $ (where $ v $ is the velocity), we can write:
$ 1 = 2axv + bv = v(2ax + b) $
This simplifies to:
$ 2ax + b = \frac{1}{v} $
Next, differentiate the expression $ 2ax + b = \frac{1}{v} $ again with respect to $ t $:
$ 2a \frac{dx}{dt} + 0 = -\frac{1}{v^2} \frac{dv}{dt} $
Using $ \frac{dx}{dt} = v $, we get:
$ 2av = -\frac{1}{v^2} \frac{dv}{dt} $
Solving for $ \frac{dv}{dt} $:
$ \frac{dv}{dt} = -2a v^3 $
Since acceleration $ f = \frac{dv}{dt} $, we have:
$ f = -2av^3 $

Answer

2bv3

Answer

-2abv2

Answer

2av2

Answer

-2av3

Motion in 1D

The relation between time t and distance x is t = ax2 + bx where a and b are constants. The acceleration is

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