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?

Question

1234 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-17T15:16:50

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

#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#

Answer 2 · 2018-03-17T15:19:12

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#