A particle of mass m=5.00 kilograms is at rest at t=0.00 seconds. a varying force f(t)=6.00t2−4.00t+3.00 is acting on the particle between t=0.00 seconds and t=5.00 seconds. find the speed v of the particle at t=5.00 seconds?

43 m/s

Explanation:

Mass, m = 5 kg

Force, F(t) = 6t² - 4t + 3

To find the speed, we first need to get the acceleration, a.

Force is the product of mass and acceleration. It is given as:

F = ma

Therefore, acceleration is:

a(t) = F(t)/m

a(t) = (6t² - 4t + 3) / 5

a(t) = 1.2t² - 0.8t + 0.6

Acceleration is the differentiation of velocity with respect to time, t. Therefore, to get velocity, v, we integrate a(t):

a(t) = dv(t) / dt

=> v(t) = (1.2/3)t³ - (0.8/2)t² + 0.6t

v(t) = 0.4t³ - 0.4t² + 0.6t

Therefore, at time t = 5secs, velocity is:

v(5) = 0.4 * (5³) - 0.4 * (5²) + 0.6 * 5

v(5) = 50 - 10 + 3

v(5) = 43 m/s

The velocity at time, t = 5 secs is 43 m/s

3.2x10^7