Question

A 100 kg ball has a force acting on it. This force is: F(t) =.4ti -.7tj

when t = 0 seconds then x = 0 meters

initial velocity: v = 3i - 2j

Question: How do you find velocity and displacement the at 4 seconds?

I am not sure how to apply integration to this problem. Please show step by step how to solve.

Answer 1

m = mass of the ball = 100 kg

F(t) = force on the ball at any time "t" = (0.4 t) i - (0.7 t) j

v_o = initial velocity of the ball at t = 0 , = 3i - 2j

acceleration of the ball at any time "t" is given as

a(t) = F(t)/m

a(t) = ( (0.4 t) i - (0.7 t) j )/ 100

a(t) = (4 x 10^-3)t i - (7 x 10^-3)t j

we know that : a(t) = dv(t)/dt

dv(t) = a(t) dt

dv(t) = ( (4 x 10^-3)t i - (7 x 10^-3)t j) dt

taking dervative both side

v(t) - v_o =  (2 x 10^-3)t² i - (3.5 x 10^-3)t² j)

v(t) - (3i - 2j) =  (2 x 10^-3)t² i - (3.5 x 10^-3)t² j)

v(t) = (3 + (2 x 10^-3)t²) i - (2 + (3.5 x 10^-3)t²) j)

at t = 4

v(4) = (3 + (2 x 10^-3)(4)²) i - (2 + (3.5 x 10^-3)(4)²) j)

v(4) = 3.032 i - 2.056 j

we know that : v(t) = dx(t)/dt

dx(t) = v(t) dt

dx(t) = (3 + (2 x 10^-3)t²) i - (2 + (3.5 x 10^-3)t²) j) dt

x(t) - x(0) = (3t + (2 x 10^-3)(t³/3) i - (2t + (3.5 x 10^-3)(t³/3) j)

x(t) = (3t + (2 x 10^-3)(t³/3) i - (2t + (3.5 x 10^-3)(t³​/3) j) Since x(0) = 0

x(t) = (3(4) + (2 x 10^-3)((4)³/3) i - (2(4) + (3.5 x 10^-3)((4)³​/3) j)

x(t) = 12.04 i - 8.08 j