Question

Could you easily teach somebody a proof of the equations of motion? How would you do this?

Accepted Answer

Proof of Ist equation:

We know that acceleration is given as:

$a = \frac{d v}{d t}$ .

Rearranging and integrating within proper limit, we have,

$d v = a d t \Rightarrow \int_{u}^{v} d v = \int_{0}^{t} a d t \Rightarrow v - u = a \int_{0}^{t} d t \Rightarrow v - u = a \int_{0}^{t} d t \Rightarrow v - u = a t$

Proof of IInd equation:

We know that acceleration is given as:

$v = \frac{d x}{d t}$ .

Rearranging and integrating within proper limit, we have,

$d x = v d t \Rightarrow \int_{0}^{x} d x = \int_{0}^{t} v d t \Rightarrow x - 0 = \int_{0}^{t} (u + a t) d t (U \sin g I s t e q u a t i o n) \Rightarrow x = u \int_{0}^{t} d t + a \int_{0}^{t} t d t \Rightarrow x = u t + \frac{1}{2} a t^{2}$

Proof of IIIrd equation:

We know that acceleration is given as:

$a = \frac{d v}{d t}$ .

which can be written as using the chain rule:

$a = \frac{d v}{d t} = \frac{d v}{d x} \frac{d x}{d t} = v \frac{d v}{d x}$ .

Rearranging and integrating within proper limit, we have,

$v d v = a d x \Rightarrow \int_{u}^{v} v d v = \int_{0}^{x} a d x \Rightarrow \frac{1}{2} (v^{2} - u^{2}) = a \int_{0}^{x} d x \Rightarrow \frac{1}{2} (v^{2} - u^{2}) = a (x - o) \Rightarrow \frac{1}{2} (v^{2} - u^{2}) = a x \Rightarrow (v^{2} - u^{2}) = 2 a x$

David Sang and Graham Jones Solutions for Chapter: Accelerated Motion, Exercise 14: REFLECTION

David Sang Physics Solutions for Exercise - David Sang and Graham Jones Solutions for Chapter: Accelerated Motion, Exercise 14: REFLECTION

Questions from David Sang and Graham Jones Solutions for Chapter: Accelerated Motion, Exercise 14: REFLECTION with Hints & Solutions

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