Physics, asked by sament1978, 8 months ago

the initial velocity of a body moving along a straight line is 7 m s - 1 it has a uniform acceleration of 4 m s -2 the distance covered by the body in the v second of its motion is​

Answers

Answered by Anonymous
1

\bigstar EXPLANATION \bigstar

  • Correct Question

The initial velocity of a body moving along a straight line is 7 m/s it has a uniform acceleration of 4 m/s^2 the distance covered by the body in the v th second of its motion is​ ____________.

  • Given

u = 7 m/s

a = 4 m/s^2

To find the distance travelled in v th second,

We have to first find the distance travelled in v seconds and then we have to find the distance travelled in (v - 1) seconds and we have to subtract the distance travelled in (v - 1) seconds from the distance travelled in v seconds

Let the distance travelled in v seconds be S

Let the distance travelled in (v - 1) seconds be S_o

From the second equation of motion, (S = ut + \frac{at^2}{2})

The distance travelled in v seconds,

S = 7v + \frac{4v^2}{2}

S = 7v + 2v^2

The distance travelled in (v - 1) seconds,

S_o = 7(v-1) + \frac{4(v-1)^2}{2}

S_o = 7v - 7 + 2(v-1)^2

S_o = 7v - 7 + 2(v^2 + 1 - 2v)

S_o = 7v - 7 + 2v^2 + 2 - 4v

S_o = 2v^2 + 3v - 5

Therefore,

The distance travelled in v th second = S - S_o

S - S_o = 7v + 2v^2  - 2v^2 - 3v + 5

S - S_o = 4v + 5

Therefore,

Distance travelled in vth second = 4v + 5

We can also find the distance travelled in v th second using the formula

S_n = u + \frac{a(2n - 1)}{2}, n is the time at which should find the distance travelled

Therefore,

Distance travelled in vth second = 7 + \frac{4(2v - 1)}{2} = 7 + 2(2v - 1) = 4v + 5

  • Equations of motion

i) v = u + at

ii) S = ut + \frac{at^2}{2}

iii) v^2 - u^2 = 2as

iv) S_n = u + \frac{a(2n-1)}{2}

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