Grade 10Mechanics

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.

Question image for A car accelerates from rest a constant rate for f

Lavesh

7 Years agoGrade 10

3 Answers

Arun

7 Years ago

y=3x

first 10 seconds:

v=u + at (u=0)

v = 10a ( t= 10)

v2 = u2 + 2as

100a2 = 2ax

x = 50a

2nd body

now u = 10a

v = 10a + 10a = 20a (v= u + at)

v2 = u2 + 2as

400a2 = 100a2 + 2ay

solving u get y = 150a

therefore y = 3x

Eshan

7 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.

NIKHIL CHOTRANI

7 Years ago

y=3x

for first 10 seconds

v=u + at (and since u=0,started from rest)

v = 10a ( till t= 10)

v2 = u2 + 2as

10a*10a = 2ax

x = 50a

2nd journey case(next 10 sec)(or from t=10sec to t=20sec)

now u = 10a

v = 10a + a*10 = 20a (v= u + at)

v2 = u2 + 2as

20a*20a= 10a*10a + 2ay

solving u get y = 150a

therefore y = 3x

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View all Questions

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.

Lavesh

3 Answers

Arun

Eshan

NIKHIL CHOTRANI

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