Question

The position of an object is described as x = 4.0t and y= 1.5t2 - 2t + 1.0.
All factors are in SI-units.
a) What is the begining velocity of the object?
b) What is his acceleration?
c) What is the direction along the x-axe at the time t=1.0 s ?

Given answers are: 4.5 m/s in -27 degree; (0; 3.0 m/s2); +14 degree

Answer 1

dx/dt = 4
dy/dt = 3 t -2

so at t = 0
V = 4 i - 2 j
|V| = sqrt (16+9) = 5 m/s
theta = tan^-1 -2/4 = -26.6 degrees

b)
Ax = 0
Ay = 3 m/s^2

c) is a typo I think.

Answer 2

|V| = sqrt (16+4) = 4.47 m/s
theta = tan^-1 -2/4 = -26.6 degrees

Answer 3

Maybe in c you mean what is the velocity angle to the x direction at t = 1 ???
dx/dt = 4
dy/dt = 3-2 = 1
tan^-1 (1/4) = 14.03 degrees

Answer 4

Thank you very much!

Answer 5

You are welcome.