Answer:
F(t) = (-6.00 N/s^2)t^2
a(t) = (-3.00 N/s^2)t^2
Because F = ma, the acceleration function is the force function divided by mass (3.50 kg). Because the force is acting to the left, a negative has been introduced.
Take the integral of the acceleration function with the power rule for integrals. Initial velocity is 8.00 m/s
∫a(t) dt=v(t)+v1
v(t)=(-1m/s^4)*t^3+9 m/s
Setting velocity equal to zero and solving for t.
v(t)=0
t^3=9s^3
t=∛9s
=2.08 s
The integral of velocity is position. The object begins at the origin so initial position is 0
∫v(t) dt= x(t)
x(t)=(-0.25m/s^4)*t^4+(9m/s)*t
Plugging the t from step 3 into the x(t) function from step 4. This is the answer to part a.
x(2.08)=14 m
plug 3.50 s into the velocity function from step 2. Speed is the absolute value of velocity. This is the answer to part b.
v(3.5)=(1 m/s^4)(3.5 s)^3+9 m/s
= -18 m/s
speed(3.5 s)=║v(3.5)║=18 m/s
(a) The distance traveled by the box before stopping is 3.44 m.
(b) The velocity of the box at 3.5 s is 40.55 m/s.
The given parameters;
The force applied on the box is given by Newton's second law of motion;
[tex]F = ma\\\\F = \frac{mv}{t} \\\\\frac{dF}{dt} = mv\\\\F(t) = 6t^2\\\\\frac{dF}{dt} = 12 t\\\\12t = mv\\\\t = \frac{mv}{12} \\\\t = \frac{3.5 \times 8}{12} \\\\t = 2.33 \ s[/tex]
The force applied to the box is calculated as;
F = 6t²
F = 6(2.33)²
F = 32.57 N
The acceleration of the box is calculated as follows;
[tex]a = \frac{F}{m} \\\\a = \frac{32.57}{3.5} \\\\a = 9.3 \ m/s^2[/tex]
The distance traveled by the box before stopping is calculated as;
[tex]v^2 = u^2 + 2as\\\\0 = u^2 + 2as\\\\-2as = u^2\\\\- s = \frac{u^2}{2a} \\\\-s = \frac{(8)^2}{2(9.3)} \\\\-s = 3.44 \ m\\\\|s| = 3.44 \ m[/tex]
The velocity of the box at 3.5 s is calculated as;
[tex]v_f = v_0+ at\\\\v_f = 8 + (9.3 \times 3.5)\\\\v_f = 40.55 \ m/s[/tex]
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