An object with a mass of #5 kg# is traveling at #2 m/s#. If the object is accelerated by a force of #f(x) = x^2 -x +3 # over #x in [1, 9]#, where x is in meters, what is the impulse at #x = 8#?

This is a tricky question, because acceleration is given as a function of x rather the parameter t... In any case write the impulse equation: #Fdt = d(mv) = mdv# So first thing we do it convert this to be in terms of #x#. From above: # F(x) = m (dv)/dt # re-write in terms of x #F(x) = m color(red)(d^2x)/dt^2# we have a differential equation of the form: #(d^2x)/dt^2 = 1/mF(x, dx/dt)# Let #v = dx/dt;# using chain rule #(dv)/dt = (dv)/dx (dx)/dt = u (dv)/dx=F(x, v)# but #F(x, dx/dt) = f(x) in [1,9]# #u (dv)/dx = 1/m f(x)# the force depends on x only not v #u (dv)/dx = 1/m f(x) = 1/m (x^2−x+3) # rearrange and solve by integrating #int vdv = 1/m int(x^2−x+3) dx# #v^2 = 2/5 (1/3x^3 - 1/2x^2 + 3x +C) # since the object started with a velocity of 2m/s for #x<1# We expect that it was #v = 2m/s#, for #x=0# #C = 10# #v(x) = sqrt(2/5 (1/3x^3 - 1/2x^2 + 3x +10) )# #v(8) = sqrt(2/5 (1/3(8)^3 - 1/2(8)^2 + 38 +10) ) = 8.31m/s#