Question

The height of an object y is given as y = 8t – 5t 2 .
(a) What is the initial velocity of an object?
(b) Find the velocity at time t= 5 sec.
(c) Calculate acceleration of the object. Is it constant or variable?

Answer 1

know, change in position per unit time is called, velocity .

let

dt

dy

=V

y

V

y

=8−10t

now at t=0 ( initially )

V

y

=8−10×0=8

Also, x=6t

differentiate wrt time

dx/dt=6

similarly , dx/dt=V

x

=6

now,

V=V

x

i

^

+V

y

j

^

= 8

i

^

+6

j

^

magnitude of velocity =∣V∣=

8²+6²