Question

a particle moves according to the relation x=2t^2-5t+6.find average velocity in the first 3 second

Answer 1

Answer:

Vavg = 1 m/ sec

Explanation:

Vavg = disp. / time

Disp. at t = 0 sec is

x = 6m

at t = 1 sec

x = 2 - 5 + 6

= 3 m

at t = 2 sec

x = 8 - 10 + 6

= 4 m

at t = 3 sec

x = 9 m

disp. = position at t = 3 sec - position at t = 0 sec

Vavg = (9 - 6)/3

= 3/3

= 1 m/sec

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