5. An object has a constant acceleration of 40 ft/sec2, an initial velocity of -20 ft/sec, and an initial position of 10 ft. Find the position function, s(t), describing the motion of the object. (10 points)

You can solve for the velocity and position functions by integrating using the fundamental theorem of calculus:

a(t) = 40 ft/s²

v(t) = v (0) + ∫₀ᵗ a(u) du

v(t) = -20 ft/s + ∫₀ᵗ (40 ft/s²) du

v(t) = -20 ft/s + (40 ft/s²) t

s(t) = s (0) + ∫₀ᵗ v(u) du

s(t) = 10 ft + ∫₀ᵗ (-20 ft/s + (40 ft/s²) u ) du

s(t) = 10 ft + (-20 ft/s) t + 1/2 (40 ft/s²) t ²

s(t) = 10 ft - (20 ft/s) t + (20 ft/s²) t ²

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.

answer: the answer is one

step-by-step explanation:

pemdas, in this situation there is no parenthesis or exponents so you go to multiplication, 1x1=1, then division 1/1= 1 then addition 1+1=2 then subtraction 2-1=1