62.5 m
Given:
Question:
What is the minimum distance required for the car to stop?
The Process:
Newton's second law of motion: [tex]\boxed{ \ \Sigma F = ma \ }[/tex]
The car is said to experience braking force in the opposite direction to the car's movement. We will find out the amount of acceleration due to braking. In this issue, it is more precisely called deceleration.
[tex]\boxed{ \ \Sigma F = ma \ } \rightarrow \boxed{ \ a = \frac{- friction \ force}{m} \ }[/tex]
The minus sign on the frictional force indicates the opposite direction to the object motion.
[tex]\boxed{ \ a = \frac{- 5000}{1000} \ } \rightarrow \boxed{ \ a = - 5 \ m/s^2 \ }[/tex]
The minus sign emphasizes the acceleration in the form of deceleration. The car will stop after traveling a certain distance which we will find out.
Next, let us use one of the following equations of uniform motion.
[tex]\boxed{ \ v^2 = u^2 + 2ad \ }[/tex]
Let us calculate the minimum distance required for the car to stop.
0² = 25² + 2(-5)(d)
10d = 625
[tex]\boxed{d = \frac{625}{10}}[/tex]
Thus, the minimum distance required for the car to stop is 62,5 m.
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Alternative Steps
We can also use the relationship between work and the change in kinetic energy (ΔKE).
Recall that,
Let us calculate the minimum distance required for the car to stop.
[tex]\boxed{ \ W = \Delta KE \ }[/tex]
[tex]\boxed{ \ -f \times d = \frac{1}{2}m(v^2 - u^2) \ }[/tex]
The minus sign on the frictional force indicates the opposite direction to the object motion.
[tex]\boxed{ \ -5000 \times d = \frac{1}{2}(1000)(0^2 - 25^2) \ }[/tex]
[tex]\boxed{ \ -5000 \times d = (500)(-625) \ }[/tex]
[tex]\boxed{ \ d = \frac{(500)(-625)}{-5000} \ }[/tex]
[tex]\boxed{ \ d = \frac{625}{10} \ }[/tex]
Thus, the minimum distance required for the car to stop is 62,5 m.
Keywords: friction forces, brakes, travelling, the minimum distance, Newton's Second Law, work, kinetic energy, velocity, uniformly accelerated motion, acceleration