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UNO [17]
3 years ago
15

(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 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).
Physics
1 answer:
nasty-shy [4]3 years ago
8 0

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

v^2 - u^2 = 2ad

To find the acceleration of the car, a:

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

F=(1100 kg)(-2.23 m/s^2)=-2451 N

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) -1.53\cdot 10^5 N

We can use again the equation

v^2 - u^2 = 2ad

To find the acceleration of the car. This time we have

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2

So now we can find the force exerted on the car by using again Newton's second law:

F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N

As we can see, the force is much stronger than the force exerted in part a).

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3 years ago
(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly
nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

v^2 - u^2 = 2ad

To find the acceleration of the car, a:

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

F=(1100 kg)(-2.23 m/s^2)=-2451 N

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) -1.53\cdot 10^5 N

We can use again the equation

v^2 - u^2 = 2ad

To find the acceleration of the car. This time we have

u=85.0 km/h = 23.6 m/s is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2

So now we can find the force exerted on the car by using again Newton's second law:

F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N

As we can see, the force is much stronger than the force exerted in part a).

8 0
3 years ago
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