The acceleration function (in m/s2) and the initial velocity are given for a particle moving along a line. a(t) = t + 6, v(0) = 3, 0 ≤ t ≤ 11 (a) find the velocity at time t. v(t) = correct: your answer is correct. m/s (b) find the distance traveled during the given time interval.

a) V = 187 s

b) d = 2024 m

Explanation:

Data

a(t) =t + 6

vo = 3

0≤t≤11

a) Find the velocity at that time

When the time = 11

a(11) = 11 + 6 = 17 m/s²

and a = v/t, so v = at

finally v = (17 m/s²) (11s)

v= 187 m/s

b) Find the distance traveled during the given time interval

Vtotal = Vfinal - v initial

v final = 187 m/s   vinitial = 3 m/s

V total = 187 - 3 = 184 m/s

and Velocity = distance/time

then distance = velocity x time

so distance = 184 m/s x 11 s

distance = 2024 m

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this