First slide

Instantaneous acceleration

Question

The distance x travelled by a particle in time t is given by t = 2x²+3x.If 'v' is the velocity, then acceleration will be :

Moderate

A

- 4v³
B

- 3v⁴
C

4v²
D

- 3v³

Solution

t=2x²+3x
So, $\frac{dt}{dx} = 4 x + 3$
$\Rightarrow (\frac{1}{v}) = 4 x + 3$
$\Rightarrow - \frac{1}{v^{2}} \frac{dv}{dx} = 4$
$\Rightarrow \frac{dv}{dx} = - 4 v^{2}$ or $\frac{vdv}{dx} = - 4 v^{3}$
$\Rightarrow a = - 4 v^{3}$

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