Question

A body is moving according to the equation x=at+bt^2-ct^3 where x= displacement a ,b and c are constant. the acceleration of the body is

Answer 1

$v= \frac{dx}{dt} \\ v = \frac{d(at+bt^2-ct^3)}{dt} \\ v= a+2bt-3ct^2 \\ a = \frac{dv}{dt} \\ a = \frac{d( a+2bt-3ct^2 )}{dt} \\ a=2b-6ct$

Answer 2

Answer:

Acceleration of the body is $2b-6ct$

Explanation:

Given: displacement =  $x=at+bt^2-ct^3$

Velocity is the change in displacement of the body per unit change in time.

Velocity is given by the formula,  $v=\frac{dx}{dt}$

$\implies v=\frac{d}{dt}[at+bt^2-ct^3]$

$\implies v=a+2bt-3ct^2$

Acceleration is the change in velocity of the body per unit change in time.

Aceleration is given by the formula,   $a=\frac{dv}{dt} = \frac{d}{dt} (\frac{dx}{dt})$

$\implies a= \frac{d}{dt}[a+2bt-3ct^2]$

$\implies a= 0+2b-6ct$

$\implies a=2b-6ct$

Therefore, acceleration of the body is $2b-6ct$

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