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Question

A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s² and an initial velocity of vector v_i = 6.00î m/s.

(a) Find the vector position of the particle at any time t (where t is measured in seconds).

: (_____ t î + ______ t² ĵ) m

(b) Find the velocity of the particle at any time t.

( ______ î +______  t ĵ) m/s

(c) Find the coordinates of the particle at t = 1.00 s.

x = _______  m

y = _______  m

(d) Find the speed of the particle at t = 1.00 s.

: _______ m/s

Answer 1

When working with kinematics problems that involve constant acceleration, the usual process is to use the kinematic equations and to break up each vector into their respective components. Here are the kinematic equations for reference...

v=v₀+at

x=x₀+v₀t+0.5at²

v²=v₀²+2a(x-x₀)

To answer part a, we can plug in the i component and j component of velocity and acceleration to get the i and j component of position. First for the i component...

x_i=x_0i+v_0it+0.5a_it²

x_i=0+6t+0=6t m

Now we can plug in for the j components...

x_j=x_0j+v_0jt+0.5a_jt²

x_j=0+0+0.5*2*t²=t² m

Putting the two together, we can get our position vector...

x=x_ii+x_jj=6ti+1tj m

Part b will have a similar process, but we will use the first kinematic equation instead. First we start with the i component of velocity...

v_i=v_0i+a_it

v_i=6+0=6 m/s

Similarly for the j component

v_j=v_0j+a_jt

v_j=0+2t=2t m/s

Putting the two together to get our velocity vector...

v=v_ii+v_jj=6i+2tj m/s

To find the position of the particle at time t=1s, we simply plug t=1 into our position vector equation from part a...

x=6ti+1tj m

x=6(1)i+1(i)j m

x=6i+1j m

For part d, we will plug in t=1 into our velocity equation from part b. After that, we will take the magnitude of our velocity vector to find the speed...

v=6i+2tj m/s

v=6i+2(1)j m/s

v=6i+2j m/s

Now to find the speed...

|v|=sqrt(6^2+2^2)=sqrt(40)=2sqrt(10) m/s

Final Answers:

a) x=6ti+1tj m

b) v=6i+2tj m/s

c) x=6i+1j m

d) speed=2sqrt(10) m/s