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