Question

An object is moving in the x direction only and its acceleration is given by a(t) =-7 m/s^5 t^3. If its initial velocity is 17 m/s in the x direction, what is the displacement of the object, in meters, when the object stops? Include a negative sign if the displacement is in the -x direction.

Answer 1

acceleration=a=-7

==>dv/dt=-7

==>v=integration of (-7*dt)

==>v=-7*t+c

where c is a constant

at t=0,v=17

==>c=17

==>v=-7*t+17 m/s

when the object stops, v=0

==>-7*t+17=0

==>t=17/7=2.4286 seconds

as v=-7*t+17

==>dx/dt=-7*t+17

==>x=integration of (-7*t+17)*dt

==>x=-3.5*t^2+17*t+c1

displace between t=0 to t=2.4286 seconds

=(-3.5*2.4286^2+17*2.4286+c1)-(3.5*0^2+17*0+c1)

=20.643 m

so displacement of the object is 20.643 m when the object stops.