Question

position of an object is given by x=4t^2-15t+12.find out velocity and accelaration at,t=0​

Answer 1

Answer:

x=4t^2-15t+12

x=4(0)^2-15(0)+12

x=0-0+12

x=12

Answer 2

Solution :-

$\bf{\star \bf Velocity, v = \dfrac{dx}{dt}}$

$: \implies \sf v = 2 \times 4t - 15$

$: \implies \sf v = 8t - 15$

at t = 0

$: \implies \sf v = - 15 ms^{-1}$

$\bf{\star \bf Acceleration, a= \dfrac{dv}{dt}}$

$: \implies \sf a = 8 ms^{-2}$