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A car accelerates from rest a constant rate for first 10 sec and covers a distance of X. It covers distance y in next 10 sec at the same acceleration.

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2 years ago

```							y=3xfirst 10 seconds:v=u + at (u=0)v = 10a ( t= 10)v2 = u2 + 2as100a2 = 2axx = 50a 2nd bodynow u = 10av = 10a + 10a = 20a (v= u + at)v2 = u2 + 2as400a2 = 100a2 + 2aysolving u get y = 150atherefore y = 3x
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2 years ago
```							Dear student,The statements are not given in the image. Kindly let us know the same so that we could help you with the solution.Regards.
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2 years ago
```							y=3xfor first 10 secondsv=u + at (and since u=0,started from rest)v = 10a ( till t= 10)v2 = u2 + 2as10a*10a = 2axx = 50a2nd journey case(next 10 sec)(or from t=10sec to t=20sec)now u = 10av = 10a + a*10 = 20a (v= u + at)v2 = u2 + 2as20a*20a= 10a*10a + 2aysolving u get y = 150atherefore y = 3x
```
one year ago
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