Hey ... I am having some struggle understanding these kind of problems .. I have

Question

Hey ... I am having some struggle understanding these kind of problems .. I have looked online for some solution but I could not find something understandable

So, Please make sure when you answer any of these question, you provide good explaination so I can folloew you..

1-A particle's position on the x-axis is given by the function x =(t^2 6.00 t+ 5.00 )m, where t is in s

Where is the particle when vx = 4.00 m/s ?

2-A particle's velocity is described by the function vx=kt^2m/s, where vx is in m/s, t is in s, and k is a constant. The particle's position at t0=0s is x0 = -5.50 m . At t1 = 1.00 s , the particle is atx1 = 7.50 m .

Determine the value of the constant k.

3-A particle's acceleration is described by the function ax =(10 t)m/s^2, where t is in s. Its initial conditions are x 0 = 280 m and v0x=0m/s at t =0s.

What is the particle's position at that time?

Explanation / Answer

1)  The particle speed is the derivative
v = dx/dt = 2t - 6.00
so t = (v + 6.00)/2
When v = 4.00 m/s
t = (4.00 + 6.00)/2 = 5.00 s

Then x= 25-30+5 =0 m (By the way, v is a function of t, not x.)

2) x(t) = x0 + (kt^3)/3
x(1) = -5.50 + ((1^3)/3)k
7.50 = -5.50 + (1/3)k
k = (3)(7.50+5.50) = 39 m/s^3

3) a(x)=10-t
v(x)=10t-0.5*t^2+c
when t=0, v0x=0, so, c1=0
v(x)=10t-0.5*t^2

x=5*t^2-(1/6)*t^3+c
when t=0, x=280, so, c2=280

x=5*t^2-(1/6)*t^3 +280
when t=20,
x=5*20^2-(1/6)*20^3 +280
=946.67 m

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects