Question

a particle moves in a straight line such that its velocity after t seconds is (2t²-t) m/s. find the distance gone in the first 3 seconds.

Answer 1

Answer:

22.5 m

Step-by-step explanation:

Calculate initial velocity (u) when t = 0, and velocity when t = 3:

when t = 0,  v = 2(0)² - 0 = 0  ⇒  u = 0

when t = 3,  v = 2(3)² - 3 = 15  ⇒  v = 15

SUVAT:

• s = s
• u = 0
• v = 15
• a =
• t = 3

Substituting values into SUVAT formula:

      $\sf s=\dfrac12(u+v)t$

$\implies \sf s=\dfrac12(0+15)3$

$\implies \sf s=22.5 \ m$