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A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time it is 14.4 m from the top; 2.00 s after that it is 25.6 m from the top, 2.00 s later 40.0 m from the top, and 2.00 s later it is 57.6 m from the top.

Part A

What is the acceleration of the sled?

Part B

What is the speed of the sled when it passes the 14.4-m point?

Part C

How much time did it take to go from the top to the 14.4-m point?

Part D

How far did the sled go during the first second after passing the 14.4-mpoint?

Added Wed, 01 Jul '15

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Solutions

-1

let object at 14.4m after n sec

distance covered in 2 sec after dat=u(n+2 -n)+0.5a[(n+2)^2 - n^2]=2u+0.5*a*4(n+1) =2an +2a +2u=25.6-14.4=11.2 m

distance covered in 2 sec after dis=2u +0.5 a[(n+4)^2 - (n+2)^2] =2u +0.5a*4 (3+n) =2u + 6a +2an =14.4 m

distance covered in next 2 sec=2u +0.5a [(n+6)^2 - (n+4)^2]=2u+0.5a*4(5+n) =2u+10a+2an=17.6 m

solving these

a=0.8 m/sec^2 (ans)

now

b)assuming u =0

n=6 sec

so V(14.4 m) =an=0.8*6=4.8 m/sec (ans)

c)n=6 sec (ans)

d) distance covered by sled during the first second after passing the 14.4-mpoint

=0.5*0.8(7^2 -6^2) =0.4*13=5.2 m (ans)

Added Wed, 01 Jul '15

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