A train that is traveling at 335 km/h needs 1.14 km to come to a complete stop. What is the braking acceleration of the train, assuming constant acceleration?
2006-09-21 02:39:33 · 6 answers · asked by ponies in outer space 1 in Education & Reference ➔ Homework Help
Use Torricelli's equation:
vf^2 = vi^2 + 2ad
vf is 0...
0 = 335 km/h ^2 + 2a (1.14 km)
0 = 112225 + 2.28a
a = 112225 km^2/h^2 / -2.28 km
a = 49221 km/h^2 (solution)
2006-09-21 03:02:53 · answer #1 · answered by ³√carthagebrujah 6 · 0⤊ 0⤋
This is the formula for acceleration:
a = s times d
a = 335 times 1.14 km
multyply both and this is the product
a = 381.9 km
2006-09-21 02:51:17 · answer #2 · answered by juliotelehit 2 · 0⤊ 0⤋
a = s * d
a = 335 * 1.14
a = 381.9
2006-09-21 02:46:16 · answer #3 · answered by Cool Z 5 · 0⤊ 0⤋
a=335/0.034
a=9853km/h/h
2006-09-21 03:37:25 · answer #4 · answered by tonima 4 · 0⤊ 0⤋
Carthage has it right...although it is a negative acceleration
2006-09-21 03:19:34 · answer #5 · answered by Professor 3 · 0⤊ 0⤋
a= s*d so...plug it in
a= 355 * 1.14
that should be your answer
2006-09-21 02:41:27 · answer #6 · answered by flpdog3 2 · 0⤊ 0⤋