ToughSTEM
ToughSTEM
A question answer community on a mission
to share Solutions for all STEM major Problems.
Cant find a problem on ToughSTEM?
0
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?
Edit
Community
1
Comment
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)
Edit
Community
1
Comment
Close

Choose An Image
or
Get image from URL
GO
Close
Back
Close
What URL would you like to link?
GO
α
β
γ
δ
ϵ
ε
η
ϑ
λ
μ
π
ρ
σ
τ
φ
ψ
ω
Γ
Δ
Θ
Λ
Π
Σ
Φ
Ω
Copied to Clipboard

to interact with the community. (That's part of how we ensure only good content gets on ToughSTEM)
OR
OR
ToughSTEM is completely free, and its staying that way. Students pay way too much already.
Almost done!