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Colt1911 [192]
3 years ago
7

Suppose you want to design an air bag system that can protect the driver at a speed 100 km/h (60 mph) if the car hits a brick wa

ll.
Estimate how fast the air bag must inflate to effectively protect the driver. How does the use of a seat belt help the driver?
Physics
1 answer:
34kurt3 years ago
3 0

When solving question that contains equations and the use mathematical computations, It is always ideal to list the parameters given.

Now, given that:

  • the speed of the car which is the initial velocity (u) = 100 km/h before it hits the wall.
  • after hitting the wall, the final velocity will be (v) = 0 km/h

Assumptions:

  • Suppose we make an assumption that the distance travelled during the collision of the car with the brick wall (S) = 1 m
  • That the car's acceleration is also constant.

∴

For a motion under constant acceleration, we can apply the kinematic equation:

\mathsf{v^2 = u^2 + 2as}

where;

v = final velocity

u = initial velocity

a = acceleration

s = distance

From the above equation, making acceleration (a) the subject of the formula:

\mathsf{v^2 - u^2 =2as }

\mathsf{a = \dfrac{v^2 - u^2 }{2s}}

The initial velocity (u) is given in km/h, and we need to convert it to m/s as it has an effect on the unit of the acceleration.

since 1 km/h = 0.2778 m/s

100 km/h = 27.78 m/s

\mathsf{a = \dfrac{(0)^2 - (27.78)^2 }{2(1)}}

\mathsf{a = \dfrac{- 771.7284 }{2}}

a = - 385.86 m/s²

Similarly, from the kinematic equation of motion, the formula showing the relation between time, acceleration and velocity is;

v = u + at

where;

v = 0

-u = at

\mathsf{t = \dfrac{-u}{a}}

\mathsf{t = \dfrac{-27.78}{-385.86}}

t = 0.07 seconds

An airbag is designed in such a way as to prevent the driver from hitting on the steering wheel or other hard substance that could damage the part of the body. The use of the seat belt is to keep the driver in shape and in a balanced position against the expansion that occurred by the airbag during the collision on the brick wall.

Thus, we can conclude that the airbag must be inflated at 0.07 seconds faster before the collision to effectively protect the driver.

Learn more about the kinematic equation here:

brainly.com/question/11298125?referrer=searchResults

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The conclusions that are specifically supported by the data in Table 1 is that An increase in the number of rubber bands causes an increase in the acceleration. That is option D.

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Acceleration is defined as the rate at which the velocity of a moving object changes with respect to time which is measured in meter per second per second (m/s²).

From the table given,

Trial 1 ----> 1 band = 0.24m/s²

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A car is up on a hydraulic lift at a garage. The wheels are free to rotate, and the drive wheels are rotating with a constant an
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Answer:

Explanation:

Given

Wheels are rotating with constant angular velocity let say \omega

Presence of constant angular velocity show that there is no angular acceleration thus there is no tangential acceleration.

But any particle on the rim will experience a constant acceleration towards center called centripetal acceleration.

(a) yes, there will be tangential velocity which is given by

v=r\cdot \omega

where r=radial distance from center

(b)tangential acceleration

there would be no tangential acceleration as velocity is constant

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7 0
3 years ago
The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9
kirill [66]

Answer:

Speed of the car 1 =V_1=8.98m/s

Speed of the car 2 =V_2=17.96m/s

Explanation:

Given:

Mass of the car 1 , M₁ = Twice the mass of car 2(M₂)

mathematically,

M₁ = 2M₂

Kinetic Energy of the car 1 = Half the kinetic energy of the car 2

KE₁ = 0.5 KE₂

Now, the kinetic energy for a body is given as

KE =\frac{1}{2}mv^2

where,

m = mass of the body

v = velocity of the body

thus,

\frac{1}{2}M_1V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

\frac{1}{2}2M_2V_1^2=0.5\times \frac{1}{2}M_2V_2^2

or

2M_2V_1^2=0.5\times M_2V_2^2

or

2V_1^2=0.5\times V_2^2

or

4V_1^2= V_2^2

or

2V_1= V_2  .................(1)

also,

\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2

or

\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2

or

2(V_1+9.0)^2=(2V_1+9.0)^2

or

\sqrt{2}(V_1+9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)

or

(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)

or

(2V_1-\sqrt{2}V_1)=(12.72-9.0)

or

(0.404V_1)=(3.72)

or

V_1=8.98m/s

and, from equation (1)

V_2=2\times 8.98m/s = 17.96m/s

Hence,

Speed of car 1 =V_1=8.98m/s

Speed of car 2 =V_2=17.96m/s

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