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Sphinxa [80]
2 years ago
5

A car advertisement states that a car can accelerate at 2 m/sec. The car starts from rest and accelerates to 70 km/hr. How much

time does it take to accelerate to 70 km/hr.​
Physics
1 answer:
Reptile [31]2 years ago
7 0

Answer:

9.72s

Explanation:

\pink{\frak{Given}}\begin{cases}\textsf{ A car starts from rest.}\\\textsf{It has an acceleration of 2m/s$^2$.}\\\textsf{ The final velocity of the car is 70km/hr.}\end{cases}

Here we need to find out the time that the car took to achieve the velocity of 70km/hr after starting from rest . Firstly convert the final velocity in m/s by multiplying it by 5/18 , as ,

\longrightarrow\sf 70km/hr = 70 \times \dfrac{5}{18} m/s =\bf 19.44\  m/s

  • As we know that the rate of change of velocity is known as acceleration , so ;

\longrightarrow\sf a =\dfrac{v - u}{t}

Plug in the respective values ,

\longrightarrow\sf 2m/s^2 = \dfrac{ 19.4 m/s -0m/s}{t} \\

Simplify ,

\longrightarrow\sf 2m/s^2 =\dfrac{19.4m/s}{t}

Cross multiply ,

\longrightarrow\sf t = \dfrac{19.44m/s}{2m/s^2}

Simplify

\longrightarrow\sf \boxed{\bf t = 9.72 s}

<h3>Hence the required answer is 9.72s.</h3>
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Answer:

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Explanation:

The frozen rock experiments a free fall, which is a type of uniform accelerated motion due to gravity and air viscosity and earth's rotation effect are neglected. In this case, we need to find the final velocity (v), measured in meters per second, of the frozen rock at given instant and whose kinematic formula is:

v = v_{o} + g\cdot t (Eq. 1)

Where:

v_{o} - Initial velocity, measured in meters per second.

g - Gravity acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we get that v_{o} = 0\,\frac{m}{s}, g = -9.807\,\frac{m}{s^{2}} and 1.5\,s, then final velocity is:

v = 0\,\frac{m}{s}+\left(-9.807\,\frac{m}{s^{2}} \right) \cdot (1.5\,s)

v = -14.711\,\frac{m}{s}

The velocity of the frozen rock at t = 1.5\,s is -14.711 meters per second.

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Explanation:

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The formula for the time period of spring is given by :

T=2\pi \sqrt{\dfrac{m}{k}}

Where

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