If the position of aparticle is given by x=5.0-9.8t+6.4t^2, what is the velocity and acceleration of the particle at t=4.0s?

2 Answers
Mar 17, 2018

v(4) = 41.4 \text( m/s)
a(4) = 12.8 \text( m/s)^2

Explanation:

x(t) = 5.0 - 9.8t + 6.4t^2 \text( m)

v(t) = (dx(t))/(dt) = -9.8 + 12.8t\text( m/s)

a(t) = (dv(t))/(dt) = 12.8\text( m/s)^2

At t = 4:

v(4) = -9.8+12.8(4) = 41.4 \text( m/s)
a(4) = 12.8 \text( m/s)^2

Mar 17, 2018

The given equation can be compared with s=ut +1/2 at^2

which is an equation of position-time relationship of a particle moving with constant acceleration.

So,rearranging the given equation,we get,

x=5-9.8*t +1/2 *12.8 t^2 (also see, at t=0,x=5)

So,the acceleration of the particle is constant i.e 12.8 ms^-2 and initial velocity u=-9.8 ms^-1

Now,we can use the equation, v=u +at to find velocity after 4s

So, v=-9.8 +12.8*4=41.4 ms^-1