Accelerated Physics Homework Help

AI Thread Summary
A boy sledding down a hill accelerates at 1.45 m/s² from rest to reach a speed of 7.60 m/s. The relevant equation used for the calculation is (vf)² = (vo)² + 2ax, where vf is the final velocity, vo is the initial velocity, and a is acceleration. The initial calculation mistakenly stated that 7.6² equals 15.2, but the correct value is 57.76. This error affects the subsequent computation of distance, which should be recalculated using the correct value. The accurate distance to reach the speed of 7.60 m/s needs to be determined based on the corrected equation.
Leo34005
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Homework Statement



A boy sledding down a hill accelerates at 1.45 m/s2. If he started from rest, in what distance would he reach a speed of 7.60 m/s?

Homework Equations



(vf)^2 = (vo)^2 + 2ax
where vf = final velocity = 7.6 m/s
vo = initial velocity = 0 m/s
a = 1.45 m/s2

x = distance = unknown

The Attempt at a Solution



a = 1.45 m/s
Vf = 7.60 m/s
Vo = 0 m/s

(7.60)^2 = (0)^2+2(1.45)(x)
15.2 = 0 + 2.9x
15.2 = 2.9x
5.143 = x
 
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You made a computation error:

7.6^2 is not 15.2

7.6^2=7.6 X 7.6 = 57.76
 
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