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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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Answer:

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Explanation:

If you cannot visualize it, just assume that the distance from station A to B is 420km. Each half is 210km.

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A ball has a mass of 1.5kg and is thrown straight up with a speed of 60m/s, what is the ball’s momentum:
madam [21]

Answer:

Assumption: the air resistance on this ball is negligible. Take g = 10\; \rm m \cdot s^{-2}.

a. The momentum of the ball would be approximately 60\;\rm kg \cdot m \cdot s^{-1} two seconds after it is tossed into the air.

b. The momentum of the ball would be approximately \rm \left(-45\; \rm kg \cdot m \cdot s^{-1}\right) three seconds after it reaches the highest point (assuming that it didn't hit the ground.) This momentum is smaller than zero because it points downwards.

Explanation:

The momentum p of an object is equal its mass m times its velocity v. That is: \vec{p} = m \cdot \vec{v}.

Assume that the air resistance on this ball is negligible. If that's the case, then the ball would accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. In other words, its velocity would become approximately 10\; \rm m \cdot s^{-1} more negative every second.

The initial velocity of the ball is 60\; \rm m \cdot s^{-1}. After two seconds, its velocity would have become 60\;\rm m \cdot s^{-1} + 2\; \rm s \times \left(-10\;\rm m \cdot s^{-1}\right) = 40\; \rm m \cdot s^{-1}. The momentum of the ball at that time would be around p = m \cdot v \approx 60\; \rm kg \cdot m \cdot s^{-1}.

When the ball is at the highest point of its trajectory, the velocity of the ball would be zero. However, the ball would continue to accelerate downwards towards the ground at a constant g \approx -10\; \rm m \cdot s^{-2}. That's how the ball's velocity becomes negative.

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liq [111]

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Therefore the momentum is 13500kg\cdot m/s

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