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3 years ago

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Suppose a car is traveling at 18.6 m/s, and the driver sees a traffic light turn red. After 0.500 s has elapsed (the reaction ti

me), the driver applies the brakes, and the car decelerates at 4.00 m/s2. What is the stopping distance of the car, as measured from the point where the driver first notices the red light

1 answer:

beks73 [17]3 years ago

4 0

Answer:

s = 52.545 m

Explanation:

First, we calculate the distance covered during the 0.5 s when the driver notices the light and applies the brake.

$s_{1} = vt\\$

where,

s₁ = distance covered between noticing light and applying brake = ?

v = speed = 18.6 m/s

t = time = 0.5 s

Therefore,

$s_{1} = (18.6\ m/s)(0.5\ s)\\s_{1} = 9.3\ m\\$

Now, we calculate the distance for the car to stop after the application of brakes. For that we use 3rd equation of motion:

$2as_{2} = V_{f}^{2} - V_{i}^{2}\\\\$

where,

s₂ = distance covered after applying brake = ?

a = deceleration = - 4 m/s²

Vf = final speed = 0 m/s

Vi = initial speed = 18.6 m/s

Therefore,

$2(- 4\ m/s^{2})s_{2} = (0\ m/s)^{2} - (18.6\ m/s)^{2}\\\\s_{2} = \frac{(18.6\ m/s)^{2})}{8\ m/s^{2}}\\\\s_{2} = 43.245\ m$

So the total distance covered by the car before stopping is:

$s = s_{1} + s_{2}\\s = 9.3\ m + 43.245\ m\\$

<u>s = 52.545 m</u>

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3) A driver in a 1000-kg car traveling at 24 m/s slams on the brakes and skids to a stop. If the coefticient of friction between

Natali5045456 [20]

Answer:

A) 37 m

Explanation:

The car is moving of uniformly accelerated motion, so the distance it covers can be calculated by using the following SUVAT equation:

$v^2 -u^2 = 2ad$ (1)

where

v = 0 m/s is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d is the length of the skid

We need to find the acceleration first. We know that the force responsible for the (de)celeration is the force of friction, so:

$F=ma=-\mu mg$

where

m = 1000 kg is the mass of the car

$\mu = 0.80$ is the coefficient of friction

a is the deceleration of the car

g = 9.8 m/s^2 is the acceleration due to gravity

The negative sign is due to the fact that the force of friction is against the motion of the car, so the sign of the acceleration will be negative because the car is slowing down. From this equation, we find:

$a=-\mu g$

And we can substitute it into eq.(1) to find d:

$v^2 -u^2 = 2(\-mu g) d\\d= \frac{v^2-u^2}{-2 \mu g}=\frac{0-(24 m/s)^2}{-2(0.80)(9.8 m/s^2)}=36.7 m \sim 37 m$

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4 years ago

A uniform cubical crate is 0.740 m on each side and weighs 600 N. It rests on the floor with one edge against a very small, fixe

Sindrei [870]

Answer:

H = 0.673

Explanation:

given,

side of cubical crate = 0.74

weight of the crate = 600 N

magnitude of force = 330 N

the Horizontal distance of its Center of mass

= 0.74/2

= 0.37

Let the required Height be H

By Balancing the Torques, we get

H x 330 N = 0.37 x 600

330 H = 222

H = 0.673

hence, the height above the floor where force is acting is equal to 0.673 m

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3 years ago

A record is dropped vertically onto a freely rotating (undriven) turntable. Frictional forces act to bring the record and turnta

alexgriva [62]

Answer:

The loss of initial Kinetic energy = 37.88 %

Explanation:

Given:

Rotational inertia of the turntable = $I_t$

Rotational inertia ( $I_r$ ) of the record = $0.61\times I_t$

According to the question:

<em>Frictional forces act to bring the record and turntable to a common angular speed.</em>

So,angular momentum will be conserved as it is an inelastic collision.

Considering the initial and final angular velocity of the turn table as $\omega _i\ ,\ \omega_f$ respectively.

Note :

Angular momentum $(L)$ = Product of moment of inertia $(I)$ and angular velocity $(\omega)$ .

Lets say,

⇒ initial angular momentum = final angular momentum

⇒ $L_i=L_f$

⇒ $(I_t)\times \omega_i = (I_t+I_r)\times \omega_f$

⇒ $\omega _f=\frac{I_t}{I_t+I_r} \times (\omega_i)$ ...equation (i)

Now we will find the ratio of the Kinetic energies.

⇒ $K_i=\frac{I_t\times \omega_i^2}{2}$ ⇒ $K_f=\frac{(I_r+I_t)\times \omega_f^2}{2}$

Their ratios:

⇒ $\frac{K_f}{K_i} =\frac{\frac{(I_t+I_r)\times \omega_f^2}{2} }{\frac{I_t\times \omega_i^2}{2} }$

⇒ $\frac{K_f}{K_i} = {\frac{(I_t+I_r)\times \omega_f^2}{2} } \times {\frac{2}{I_t\times \omega_i^2}}$

Plugging the values of $\omega _f^2$ as $\omega _f^2 =(\frac{I_t}{I_t+I_r} \times \omega_i\ )^2$ from equation (i) in the ratios of the Kinetic energies.

⇒ $\frac{K_f}{K_i} =\frac{(I_t+I_r)\times \frac{(I_t)^2}{(I_t+I_r)^2} \times \omega_i^2}{I_t\times \omega_i^2} =\frac{(I_t)^2}{(I_t+I_r)}\times \frac{1}{I_t}=\frac{I_t}{I_t+I_r}$

Now,

The Kinetic energy lost in fraction can be written as:

⇒ $\frac{K_f-K_i}{K_i}$

Now re-arranging the terms.

$\frac{K_f-K_i}{K_i} =(\frac{K_f}{K_i} -1)= \frac{I_t}{I_t+I_r} -1=\frac{I_t-I_t-I_r}{I_t+I_r} =\frac{-I_r}{(I_t+I_r)}$

Plugging the values of $I_r$ and $I_t$ .

⇒ $\frac{K_f}{K_i} = \frac{-0.61I_t}{0.61I_t+I_t} =\frac{-0.61}{1.61} =-0.3788$

To find the percentage we have to multiply it with $100$ and here negative means for loss of Kinetic energy.

⇒ $\frac{K_f}{K_i} = =-0.3788\times 100= 37.88$

So the percentage of the initial Kinetic energy lost is 37.88

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