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2 years ago

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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?

1 answer:

hoa [83]2 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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3 years ago

A 13561 N car traveling at 51.1 km/h rounds

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a) The centripetal acceleration of the car is 0.68 m/s²

b) The force that maintains circular motion is 940.03 N.

c) The minimum coefficient of static friction between the tires and the road is 0.069.

Explanation:

a) The centripetal acceleration of the car can be found using the following equation:

$a_{c} = \frac{v^{2}}{r}$

Where:

v: is the velocity of the car = 51.1 km/h

r: is the radius = 2.95x10² m

$a_{c} = \frac{(51.1 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2.95 \cdot 10^{2} m} = 0.68 m/s^{2}$

Hence, the centripetal acceleration of the car is 0.68 m/s².

b) The force that maintains circular motion is the centripetal force:

$F_{c} = ma_{c}$

Where:

m: is the mass of the car

The mass is given by:

$P = m*g$

Where P is the weight of the car = 13561 N

$m = \frac{P}{g} = \frac{13561 N}{9.81 m/s^{2}} = 1382.4 kg$

Now, the centripetal force is:

$F_{c} = ma_{c} = 1382.4 kg*0.68 m/s^{2} = 940.03 N$

Then, the force that maintains circular motion is 940.03 N.

c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:

$F_{c} = F_{\mu}$

$F_{c} = \mu N = \mu P$

$\mu = \frac{F_{c}}{P} = \frac{940.03 N}{13561 N} = 0.069$

Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.

I hope it helps you!

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