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Vaselesa [24]
3 years ago
9

An initially motionless test car is accelerated to 115 km/h in 8.58 s before striking a simulated deer. The car is in contact wi

th the faux fawn for 0.815 s, after which the car is measured to be traveling at 60.0 km/h. What is the magnitude of the acceleration of the care before the collision? What is the magnitude of the average acceleration of the car during the collision?
Physics
1 answer:
hoa [83]3 years ago
4 0

Answer:

a)       a = 3.72 m / s², b)    a = -18.75 m / s²

Explanation:

a) Let's use kinematics to find the acceleration before the collision

             v = v₀ + at

as part of rest the v₀ = 0

             a = v / t

Let's reduce the magnitudes to the SI system

              v = 115 km / h (1000 m / 1km) (1h / 3600s)

              v = 31.94 m / s

              v₂ = 60 km / h = 16.66 m / s

l

et's calculate

             a = 31.94 / 8.58

             a = 3.72 m / s²

b) For the operational average during the collision let's use the relationship between momentum and momentum

            I = Δp

            F Δt = m v_f - m v₀

            F = \frac{m ( v_f - v_o)}{t}

            F = m [16.66 - 31.94] / 0.815

            F = m (-18.75)

Having the force let's use Newton's second law

            F = m a

            -18.75 m = m a

             a = -18.75 m / s²

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1.
Gwar [14]

<u>Answer:</u>

For 1: The correct option is Option C.

For 3: The final velocity of the opponent is 1m/s

<u>Explanation: </u>

During collision, the energy and momentum remains conserved. The equation for the conservation of momentum follows:

m_1u_1+m_2u_2=m_1v_1+m_2v_2      ...(1)

where,

m_1,u_1\text{ and }v_1 are the mass, initial velocity and final velocity of first object

m_2,u_2\text{ and }v_2 are the mass, initial velocity and final velocity of second object

<u>For 1:</u>

We are Given:

m_1=150g=0.15kg\\u_1=?m/s\\v_1=0.85m/s\\m_2=3500g=3.5kg\\u_2=0m/s\\v_2=0.85m/s

Putting values in equation 1, we get:

(0.15\times u_1)+(3.5\times 0)=(3.5+0.15)\times 0.85\\\\u_1=20.683\approx 21m/s

Hence, the correct answer is Option C.

  • <u>For 2: </u>

Impulse is defined as the product of force applied on an object and time taken by the object.

Mathematically,

J=F\times t

where,

F = force applied on the object

t = time taken

J = impulse on that object

Impulse depends only on the force and time taken by the object and not dependent on the surface which is stopping the object.

Hence, the impulse remains the same.

  • <u>For 3:</u>

Let the speed in right direction be positive and left direction be negative.

We are Given:

m_1=240kg\\u_1=0m/s\\v_1=-1m/s\\m_2=80kg\\u_2=-2m/s\\v_2=?m/s

Putting values in equation 1, we get:

(240\times 0)+(80\times (-2))=(240\times (-1))+(80\times v_2)\\\\v_2=1m/s

Hence, the final velocity of the opponent is 1m/s and has moved backwards to its direction of the initial velocity.

4 0
3 years ago
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