Question

the position of an object is given by the equation x = 3.0t^2+1.5t+4.5, where x in meters and t is in seconds. what is the instantaneous acceleration of the object at t = 3.0 s?

Answer 1

Answer:

6 m/s²

Explanation:

$x(t) = 3t^2+1.5t+4.5\\v(t)=\frac{dx}{dt}=6t + 1.5\\a(t)=\frac{d^2x}{dt^2}= 6$

Answer 2

Instantaneous acceleration of the object at t = 3.0 s will be 6 m/$s^{2}$

What is meant by instantaneous acceleration ?

Instantaneous acceleration is defined as the ratio of change in velocity during a given time interval such that the time interval goes to zero

x = 3$t^{2}$ + 1.5 t + 4.5

v = dx/dt = d(3$t^{2}$ + 1.5 t + 4.5) /dt = 6t + 1.5

a = dv/dt = $d^{2} x$/$dt^{2}$ = d(6t + 1.5 ) / dt = 6 m/$s^{2}$

instantaneous acceleration of the object at t = 3.0 s will be 6 m/$s^{2}$

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