acceleration Q, thanks (1 Viewer)

Drdusk · Oct 13, 2019

shmitler said:
View attachment 27311

$\text{We know that}\hspace{2mm}a = v\dfrac{dv}{dx}\Rightarrow a = v\times 2x =2x(x^2+2)$

$\text{When x = 1}\hspace{2mm}a = 2(1+2) = 6 \hspace{2mm}m/s$

shmitler · Oct 13, 2019

Thanks

howyagoing12 · Oct 23, 2019

Drdusk said:
$\text{We know that}\hspace{2mm}a = v\dfrac{dv}{dx}\Rightarrow a = v\times 2x =2x(x^2+2)$

$\text{When x = 1}\hspace{2mm}a = 2(1+2) = 6 \hspace{2mm}m/s$

Why isn't acceleration just 2x?

fan96 · Oct 23, 2019

howyagoing12 said:
Why isn't acceleration just 2x?

Because your velocity is given in terms of $x$ , not $t$ .

The equation is

$a = \frac{dv}{dt},$

so if your $v$ is not in terms of $t$ , you can't just differentiate in that way.

acceleration Q, thanks (1 Viewer)

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