Physics, asked by kathusanthosh2002, 10 months ago

a car starts from rest and moves with constant acceleration .the ratio of distance covered by the car in nth second to that covered in n second is

Answers

Answered by Anonymous
82

Given :

  • Initial velocity (u) = 0 m/s
  • Acceleration = a (constant)

To Find :

  • Ratio of distance covered in nth second to that distance covered in n seconds.

Solution :

We're given that initial velocity is 0 m/s as the car starts from rest and accelerates constantly. So,

Use formula for distance travelled in nth second

\longrightarrow \sf{S_{n} \: = \: u \: + \: \dfrac{1}{2} a(2n\ -\ 1)} \\ \\ \longrightarrow \sf{S_n\ =\ 0\ +\ \dfrac{1}{2} a(2n\ -\ 1)} \\ \\ \longrightarrow \sf{S_n\ =\ \dfrac{1}{2} a(2n\ -\ 1) \: \: \: \: \: \: \: \: \: ...(1)}

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Now, use 2nd equation of motion :

\longrightarrow \sf{S\ =\ ut\ +\ \dfrac{1}{2} at^2} \\ \\ \longrightarrow \sf{S\ =\ 0(n)\ +\ \dfrac{1}{2} an^2} \\ \\ \longrightarrow \sf{S\ =\ \dfrac{1}{2} an^2 \: \: \: \: \: \: \: \: \: \: ...(2)}

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Dividing 2 by 1

\longrightarrow \sf{\dfrac{S_n}{S}\ =\ \dfrac{\cancel{\dfrac{1}{2}} a(2n\ -\ 1)}{\cancel{\dfrac{1}{2}}(an^2)}} \\ \\ \longrightarrow \sf{\dfrac{S_n}{S}\ =\ \dfrac{\cancel{a}(2n\ -\ 1)}{\cancel{a}n^2}} \\ \\ \longrightarrow \sf{\dfrac{S_n}{S}\ =\ \dfrac{2n\ -\ 1}{n^2}} \\ \\ \longrightarrow \sf{Ratio\ =\ 2n\ -\ 1\ :\ n^2}

Answered by EliteSoul
54

GIVEN :

  • Car starts from rest
  • Acceleration = Constant

TO FIND :

  • RATIO OF DISTANCE COVERED IN nth SECOND AND TO THAT COVERED IN n SECOND.

SOLUTION :

Here initial speed (u) = 0

Acceleration (a) = a m/s²

We know that,

Distance covered in nth second :

S(nth) = u + ½ a(2n - 1)

→ S(nth) = 0 + ½ a (2n - 1)

S(nth) = ½ a(2n - 1). ...(i)

Now using 2nd equation of motion to find distance covered in n seconds :

S(n) = ut + ½ at²

→ S(n) = 0 × n + ½ a × n²

S(n) = ½ an². ....(ii)

Now finding ratio :

→ S(nth) : S(n) = ½a(2n - 1) : (½ an²)

→ S(nth) : S(n) = ½ a(2n - 1)/ ½ an²

S(nth) : S(n) = (2n - 1) :

Ratio of distance covered in nth second to that of distance covered in n second = (2n - 1) :

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