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Vika [28.1K]
3 years ago
10

in an effort to reach the store before it closed the motorist increase the speed of his car from 20 m per second to 60 meters pe

r second over a four second. What was the car's acceleration ​
Physics
1 answer:
Ber [7]3 years ago
7 0

Answer:

10

Explanation:

Givens

vi = 20 m/s

vf = 60 m/s

t = 4 second.

Formula

a = (vf - vi) / t

a = (60 - 20)/4

a = 40 / 4

a = 10 m/s^2

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Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

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⇒\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})

⇒\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}

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Applying logarithm on both sides,

⇒log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}

⇒log(2)=\frac{2}{\frac{L}{R}}

⇒\frac{L}{R}=\frac{2}{log2}

Now substitute i(t)=\frac{3}{4}imax=\frac{3E}{4R}

⇒\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})

⇒\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}

⇒\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}

⇒log(4)=\frac{t}{\frac{L}{R}}

⇒t=log4\frac{L}{R}

now subs. \frac{L}{R}=\frac{2}{log2}

⇒t=log4\frac{2}{log2}

also log4=log2^{2}=2log2

⇒t=2log2\frac{2}{log2}

⇒t=4

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