Suppose that a 1000 kg car is traveling at 25 m/s (aprox 55mph). Its brakes can apply a force of 5000N. What is the minimum distance required for the car to stop?

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]

v = final velocity → v = 0 m/s
u = initial velocity → u = 25 m/s
a = - 5 m/s²
d = distance travelled

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,

work done (W) = force x distance,
[tex]kinetic \ energy = \frac{1}{2}mv^2[/tex]
and W = ΔKE.

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.

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Keywords: friction forces, brakes, travelling, the minimum distance, Newton's Second Law, work, kinetic energy, velocity, uniformly accelerated motion, acceleration