# (a) Calculate the force needed to bring a 900 kg car to rest from a speed of 85.0 km/h in a...

## Question:

(a) Calculate the force needed to bring a 900 kg car to rest from a speed of 85.0 km/h in a distance of 105 m (a fairly typical distance for a non-panic stop).

(b) Suppose instead the car hits a concrete abutment at full speed and is brought to a stop in 2.00 m. Calculate the force exerted on the car and compare it with the force found in part (a), i.e. find the ratio of the force in part(b) to the force in part (a).

## Application of Newton's Second Law of Motion

This law states the relationship between the force applied to any object and the acceleration produced by it on the object.

This law also tells the variation of momentum change of any particle per in unit time due to the application of a force.

Note: When a particle moves with a constant acceleration then three equations of motion are valid whose one part has been used in the answer and explanation section.

• m = mass of the car = 900 kg
• u = initial speed of the car = 85 km/h = 23.61 m/s
• v = final velocity of the car = 0 m/s
• a = constant acceleration of the car
• s = distance traveled by the car
• F = force needed to stop the car

Part (a):

In this part,

u = 23.61 m/s

v = 0 m/s

s = distance travelled by car to stop = 105 m.

Let us first calculate the value of 'a'.

{eq}v^2 = u^2 +2as\\ \Rightarrow a = \dfrac{v^2-u^2}{2s}\\ \Rightarrow a = \dfrac{(0\ m/s)^2-(23.61\ m/s)^2}{2(105\ m)}\\ \Rightarrow a = -2.65\ m/s^2\\ \therefore F = ma = 900\ kg \times (-2.65\ m/s^2) = -2385\ N {/eq}

Hence a force of {eq}F_a = 2385\ N {/eq} is required to stop the car in this part of the question.

Part (b):

In this part, the car hits a concrete abutment to stop by traveling a distance of 2 m.

u = 23.61 m/s

v = 0 m/s

s = distance traveled by car before stopping = 2 m.

Let us first calculate the value of 'a'.

{eq}v^2 = u^2 +2as\\ \Rightarrow a = \dfrac{v^2-u^2}{2s}\\ \Rightarrow a = \dfrac{(0\ m/s)^2-(23.61\ m/s)^2}{2(2\ m)}\\ \Rightarrow a = -139.35\ m/s^2\\ \therefore F = ma = 900\ kg \times (-139.35\ m/s^2) = -125415\ N {/eq}

Hence a force of {eq}F_b = 125415\ N {/eq} is required to stop the car in this part of the question.

Now,

{eq}\dfrac{F_b}{F_a} = \dfrac{125415\ N}{2385\ N} = 52.58 {/eq}

This means that the force applied (in part (b)) by the concrete abutment is 52.58 times greater than the force applied (in part (a)) to stop the car.