Can someone pretty please help? will give brainliest and thats a promise
2 answers:
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Answer
given,
position of particle
x(t)= A t + B t²
A = -3.5 m/s
B = 3.9 m/s²
t = 3 s
a) x(t)= -3.5 t + 3.9 t²
velocity of the particle is equal to the differentiation of position w.r.t. time.
------(1)
velocity of the particle at t = 3 s
v = -3.5 + 7.8 x 3
v = 19.9 m/s
b) velocity of the particle at origin
time at which particle is at origin
x(t)= -3.5 t + 3.9 t²
0 = t (-3.5 + 3.9 t )
t = 0,
t = 0 , 0.897 s
speed of the particle at t = 0.897 s
from equation (1)
v = -3.9 + 7.8 t
v = -3.9 + 7.8 x 0.897
v = 3.1 m/s