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Grade: 10
        
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.
one year ago

Answers : (3)

Arun
22583 Points
							

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

one year ago
Eshan
askIITians Faculty
2095 Points
							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.
one year ago
NIKHIL CHOTRANI
32 Points
							

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

7 months ago
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