An express train moving at 30m/s reduces its speed to 10 m/s in a distance of 240 m.If the breaking force is increased by 12.5% in the beginning find the distance that it travels before coming to rest
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The equation of motion to use is :
V² = U² + 2as
V = 10m/s U = 30m / s S= 240m
Doing the substitution we have :
100 = 900 + 480a
- 800 = 480a
a = - 5/3 m/s²
Basic definition of force
F = ma
Thus a = F/m = -5/3 m/s²
a is directly proportional to F.
An increase in F by x increases the acceleration by x also.
Since our breaking force is increased 12.5% then the new acceleration is:
a (new) = (1 + 0.125) × -5/3 m/s² = -15/8 m/s²
Using the same original equation of motion and solve for S
S = (V² - U²) / (2a)
Substituting:
this time V = 0 m/s (the train comes to rest)
S = ((0) - (900) / (2 × -15/8 m/s²) = - 900 ÷ - 15/8
900 × 8 / 15 = 240 m
V² = U² + 2as
V = 10m/s U = 30m / s S= 240m
Doing the substitution we have :
100 = 900 + 480a
- 800 = 480a
a = - 5/3 m/s²
Basic definition of force
F = ma
Thus a = F/m = -5/3 m/s²
a is directly proportional to F.
An increase in F by x increases the acceleration by x also.
Since our breaking force is increased 12.5% then the new acceleration is:
a (new) = (1 + 0.125) × -5/3 m/s² = -15/8 m/s²
Using the same original equation of motion and solve for S
S = (V² - U²) / (2a)
Substituting:
this time V = 0 m/s (the train comes to rest)
S = ((0) - (900) / (2 × -15/8 m/s²) = - 900 ÷ - 15/8
900 × 8 / 15 = 240 m
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