An object that is accelerating may be

1 answer:

8 0

The answer of this question is D. All of the above

Acceleration happen when an obeject change its velocity. It has nothing to do with speed.

The huge misconception about acceleration is when we thought it only aply if we increase our speed ( in a sport match, sportcaster often describe acceleration as an increase in players speed)

slower, faster, right , left, it does not matter, as long as that object change its velocity, it accelerates

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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 partic

olga_2 [115]

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

$a = \frac{F}{m}$ ...... (1)

acceleration (a) $= \frac{dv}{dt}$ ......(2)

substituting (2) into (1)

Hence, F $= \frac{mdv}{dt}$

$\frac{dv}{dt} = \frac{F}{m}$

$dv = \frac{F}{m} dt$

$dv = \frac{1}{m}Fdt$

Integrating both sides

$\int\limits {} \, dv = \frac{1}{m} \int\limits {F(t)} \, dt$

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

$v = \frac{1}{m} \int\limits^5_0 {F(t)} \, dt$ ......(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):

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