What are the particle's position and velocity

[brad sue]

I have no idea how to find a solution forthis exercice:

A particle moving with an initial velocity v=(30m/s)j undergoes an acceleration of
a=[3.5m/s² +(0.7m/s⁵)t³]i+[2m/s²-(0.3m/s⁴)t²]j.

what are the particle's position and velocity after 30s assuming that it starts at the origin?

i am confused by the powers 4 and 5 in a . please give me some hints to answer the problem.

Brad

 

[mukundpa]

These powers of t shows that the motion is not uniformally accelerated and so you can't use equations of motion.

integrate the acceleration to get velocity and then for position.

 

[lightgrav]

a_x gets quicker and quicker as time goes by ...
by t=2s, a_x = (3.5 + 5.6)m/s^2 , a_y = (2-1.2)m/s^2 .

v(t) = v_o + integral of a(t)dt ... x(t) = integral of v(t)dt
they're just powers, but I'd do j and i separately.

 

[gnpatterson]

The extra "powers" are the correct units of the constants that are given. the constants are part of a formula that gives an acceleration, the constant is multiplied a time (in seconds) squared or cubed, so to generate the acceleration in m/s/s the constants must be defined in m/s/s/s/s and m/s/s/s/s/s. If someone wished to do this sort of thing in a different unit of time then they would have to convert the constants into the new system of units.

but as has already been stated you really don't need to worry about them.

 

[Doc Al]

A particle moving with an initial velocity v=(30m/s)j undergoes an acceleration of
a=[3.5m/s² +(0.7m/s⁵)t³]i+[2m/s²-(0.3m/s⁴)t²]j.

Writing the units in the equation makes it look more confusing than it is. You can rewrite it like this using standard units (distance in m; time in s; acceleration in m/s^2):
a=[3.5 + 0.7t³]i+[2 - 0.3t²]j