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3 years ago

Rihanna can go from 0-60 miles per hour in 3.5 seconds. Calculate the acceleration.

Physics

1 answer:

Anton [14]3 years ago

8 0

Answer:

7.66 m/s²

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 0 mph

Final velocity (v) = 60 mph

Time (t) = 3.5 s

Acceleration (a) =.?

Next, we shall convert mile per hour (mph) to metre per second (m/s). This can be obtained as follow:

Recall:

1 mph = 0.447 m/s

Initial velocity (u) = 0 mph

Initial velocity (u) = 0 × 0.447 = 0 m/s

Final velocity (v) = 60 mph

Final velocity (v) = 60 × 0.447 = 26.82 m/s.

Finally, we shall determine the acceleration of Rihanna as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 26.82 m/s.

Time (t) = 3.5 s

Acceleration (a) =.?

a = (v – u) /t

a = (26.82 – 0) / 3.5

a = 26.82 / 3.5

a = 7.66 m/s²

Thus, the acceleration of Rihanna is 7.66 m/s²

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sergey [27]

Answer:

carbon dioxide (CO2)

Explanation:

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3 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$