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A particle of mass [tex]m = 5.00[/tex] kilograms is at rest at [tex]t = 0.00[/tex] seconds. A varying force [tex]f(t) = 6.00t^2 - 4.00t + 3.00[/tex] is acting on the particle between [tex]t = 0.00[/tex] seconds and [tex]t = 5.00[/tex] seconds.

Find the speed [tex]v[/tex] of the particle at [tex]t = 5.00[/tex] seconds. Express your answer in meters per second to three significant figures.

Answer :

Answer:

The speed v of the particle at t=5.00 seconds = 43 m/s

Explanation:

Given :

mass m = 5.00 kg

force f(t) = 6.00t2−4.00t+3.00 N

time t between t=0.00 seconds and t=5.00 seconds

From mathematical expression of Newton's second law;

Force = mass (m) x acceleration (a)

F = ma

[tex]a = \frac{F}{m}[/tex] ...... (1)

acceleration (a) [tex]= \frac{dv}{dt}[/tex] ......(2)

substituting (2) into (1)

Hence, F [tex]= \frac{mdv}{dt}[/tex]

[tex]\frac{dv}{dt} = \frac{F}{m}[/tex]

[tex]dv = \frac{F}{m} dt[/tex]

[tex]dv = \frac{1}{m}Fdt[/tex]

Integrating both sides

[tex]\int\limits {} \, dv = \frac{1}{m} \int\limits {F(t)} \, dt[/tex]

The force is acting on the particle between t=0.00 seconds and t=5.00 seconds;

[tex]v = \frac{1}{m} \int\limits^5_0 {F(t)} \, dt[/tex] ......(3)

Substituting the mass (m) =5.00 kg of the particle, equation of the varying force f(t)=6.00t2−4.00t+3.00 and calculating speed at t = 5.00seconds into (3):

[tex]v = \frac{1}{5} \int\limits^5_0 {(6t^{2} - 4t + 3)} \, dt[/tex]

[tex]v = \frac{1}{5} |\frac{6t^{3} }{3} - \frac{4t^{2} }{2} + 3t |^{5}_{0}[/tex]

[tex]v = \frac{1}{5} |(\frac{6(5)^{3} }{3} - \frac{4(5)^{2} }{2} + 3(5)) - 0|[/tex]

[tex]v = \frac{1}{5} |\frac{6(125)}{3} - \frac{4(25)}{2} + 15 |[/tex]

[tex]v = \frac{1}{5} |\frac{750}{3} - \frac{100}{2} + 15 |[/tex]

[tex]v = \frac{1}{5} | 250 - 50 + 15 |[/tex]

[tex]v = \frac{215}{5}[/tex]

v = 43 meters per second

The speed v of the particle at t=5.00 seconds = 43 m/s

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