Calculus Based Physic I

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mohsen1988 | Student

r = b t^3 i + c t j

==> v = dr/dt = 3 b t^2 i + c j

==> a = dv/dt = 6 b t i + 0 j

==> F = m a = 6 m b t i = (6*20*2.1*5.5) i = 1386 i = 1400 i

sciencesolve | Student

You need to use the algebraic form of Newton's second law to evaluate the force that acts on the `20Kg` object, under the given conditions, such that:

`barF = m*bar a`

Since the problem provides the vector that indicates the position of the object, you may evaluate the acceleration vector, such that:

`a(t) = r''(t) = x''(t)bar i + y''(t) bar j`

`r(t) = bt^3bar i + ct bar j`

You need to differentiate the position vector with respect to t to evaluate the velocity vector, such that:

`v(t) = r'(t) = 3bt^2 bar i + c bar j`

You need to differentiate the velocity vector with respect to t to evaluate the accleration vector, such that:

`r''(t) = 6bt*bar i + 0`

`a(t) = r''(t) = 6bt*bar i`

You need to evaluate the force at the moment `t = 5.5 s` , hence, replacing `b = 2.1m*s^(-3)` and `5.5 s` for` t` , yields:

`a(t) = 6*2.1m*s^(-3)*5.5 s*bar i`

`a(t) = 69.3m*s^(-2)`

You need to evaluate the force, at `t = 5.5 s` , such that:

`F = 20*69.3 bar i = 1386 bar i N`

Hence, evaluating the force, under the given conditions, yields `F = 1386 bar i N` .

aruv | Student



(i) m=?

By Newton's second law


100=m x 3

m=100/3 kg

(ii) Block is rest position


By second equation of linear motion





By equation of motion





`v=30 m//sec`

Zaca | Student

One equation for force, is F=ma.

Since we have an equation for position, we can take two derivatives to get the equation for acceleration:

Original equation:

`vecr= bt^3hati + cthatj`

First derivative (velocity):

`vecr ' = 3bt^2hati + chatj`

Second derivative (acceleration):

`vecr '' = 6bthati`

To find acceleration plug in the values for b and t:



`vecr '' = 6(2.1m/s^3)(5.5s)=69.3m/s^2`

Accel = 69.3 m/s^2

Mass = 20 kg

Force = ma = 69.3 m/s^2 * 20 kg = 1,386 N

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