Question

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?

Answer 1

Assuming the the car is travelling to the right (→):

$R(\rightarrow), F = ma$

$\implies a = \frac{F}{m}$

$= \frac{-5000}{1000}$

$= -5 \ ms^{-2}$

The information we know:

$u = 25 \ ms^{-1}$

$v = 0$

$a = -5 \ ms^{-2}$

Using one of the equations of motion:

$v^2 = u^2 + 2as$

$\implies s = \frac{v^2 -u^2}{2a}$

$= \frac{0^2 - 25^2}{2\times (-5)}$

$= \frac{625}{10}$

$= 62.5\ m$

Answer 2

62.5 m

Further explanation

Given:

• A 1000 kg car is traveling at 25 m/s (approx 55 mph).  
• Its brakes can apply a force of 5000 N.  

Question:

What is the minimum distance required for the car to stop?

The Process:

Newton's second law of motion: $\boxed{ \ \Sigma F = ma \ }$

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.

$\boxed{ \ \Sigma F = ma \ } \rightarrow \boxed{ \ a = \frac{- friction \ force}{m} \ }$

The minus sign on the frictional force indicates the opposite direction to the object motion.

$\boxed{ \ a = \frac{- 5000}{1000} \ } \rightarrow \boxed{ \ a = - 5 \ m/s^2 \ }$

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.

$\boxed{ \ v^2 = u^2 + 2ad \ }$

• 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

$\boxed{d = \frac{625}{10}}$

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,
• $kinetic \ energy = \frac{1}{2}mv^2$
• and W = ΔKE.

Let us calculate the minimum distance required for the car to stop.

$\boxed{ \ W = \Delta KE \ }$

$\boxed{ \ -f \times d = \frac{1}{2}m(v^2 - u^2) \ }$

The minus sign on the frictional force indicates the opposite direction to the object motion.

$\boxed{ \ -5000 \times d = \frac{1}{2}(1000)(0^2 - 25^2) \ }$

$\boxed{ \ -5000 \times d = (500)(-625) \ }$

$\boxed{ \ d = \frac{(500)(-625)}{-5000} \ }$

$\boxed{ \ d = \frac{625}{10} \ }$

Thus, the minimum distance required for the car to stop is 62,5 m.

Learn more

1. A case problem of uniformly accelerated motion and Newton's Second Law brainly.com/question/11181200
2. Finding the acceleration between two vectors brainly.com/question/6268248
3. Determine the flea's acceleration brainly.com/question/5424148

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