Ask question

Ask question

All categories

4 years ago

5

Li is riding her bicycle at 8.0 m/s. She slows down to 4.0 m/s. Her change in velocity is m/s. If Li takes 2 seconds to make thi

s change, her acceleration is m/s2.

2 answers:

malfutka [58]4 years ago

8 0

Her change in speed is (4 m/s - 8 m/s) = -4 m/s.

Her acceleration is

(change in speed) / (time for the change)

= ( -4 m/s ) / (2 sec)

= -2 m/s² .

forsale [732]4 years ago

4 0

You will have to use this formula:
$v = vo + a \times t$

Final Velocity (V) = 4m/s
Initial Velocity (Vo) = 8m/s
Acceleration (a) = ? m/s^2
Time (t) = 2 secs

Then:

-> 4 = 8 + a x 2
-> 4 - 8 = 2a
-> -4 = 2a
-> a = -4/2
-> a = -2 m/s^2

Ps: It's value is negative because the she was in retrograde motion.

Answer: Her acceleration is -2 m/s^2.

You might be interested in

The 18th hole at Pebble Beach Golf Course is a dogleg to the left of length 496.0 m. The fairway off the tee is taken to be the

dimulka [17.4K]

Answer:

(472i + 80.3j) m

Explanation:

Given the following :

Distance of tee shot = 300m

Distance of second shot = 189.0 m

Displacement r1 of tee shot :

r1 = 300 mi

Displacement r2 of second shot :

r2 = 172.0 mi + 80.3 mj

The final displacement of the golf ball from the tee:

r_final = r1 + r2

r_final = (300i)m + (172.0i + 80.3j) m

r_final = (300 + 172)i m + 80.3j m

r_final = (472i + 80.3j) m

7 0

3 years ago

As a result of friction, the angular speed of a wheel changes with time according to dθ/dt = ω0e^−σt where ω0 and σ are constant

NNADVOKAT [17]

Hi there!

a.

We can use the initial conditions to solve for w₀.

It is given that:

$\frac{d\theta}{dt} = w_0e^{-\sigma t}$

We are given that at t = 0, ω = 3.7 rad/sec. We can plug this into the equation:

$\omega(0)= \omega_0e^{-\sigma (0)}\\\\3.7 = \omega_0 (1)\\\\\omega_0 = \boxed{3.7 rad/sec}$

Now, we can solve for sigma using the other given condition:

$2 = 3.7e^{-\sigma (8.6)}\\\\.541 = e^{-\sigma (8.6)}\\\\ln(.541) = -\sigma (8.6)\\\\\sigma = \frac{ln(.541)}{-8.6} = \boxed{0.0714s^{-1}}$

b.

The angular acceleration is the DERIVATIVE of the angular velocity function, so:

$\alpha(t) = \frac{d\omega}{dt} = -\sigma\omega_0e^{-\sigma t}\\\\\alpha(t) = -(0.0714)(3.7)e^{-(0.0714) (3)}\\\\\alpha(t) = \boxed{-0.213 rad\sec^2}$

c.

The angular displacement is the INTEGRAL of the angular velocity function.

$\theta (t) = \int\limits^{t_2}_{t_1} {\omega(t)} \, dt\\\\\theta(t) = \int\limits^{2.5}_{0} {\omega_0e^{-\sigma t}dt\\\\$

$\theta(t) = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=2.5} \atop {t_1=0}} \right.$

$\theta = -\frac{3.7}{0.0714}e^{-0.0714 t}\left \| {{t_2=2.5} \atop {t_1=0}} \right. \\\\\theta= -\frac{3.7}{0.0714}e^{-0.0714 (2.5)} + \frac{3.7}{0.0714}e^{-0.0714 (0)}$

$\theta = 8.471 rad$

Convert this to rev:

$8.471 rad * \frac{1 rev}{2\pi rad} = \boxed{1.348 rev}$

d.

We can begin by solving for the time necessary for the angular speed to reach 0 rad/sec.

$0 = 3.7e^{-0.0714t}\\\\t = \infty$

Evaluate the improper integral:

$\theta = \int\limits^{\infty}_{0} {\omega_0e^{-\sigma t}dt\\\\$

$\lim_{a \to \infty} \theta = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=a} \atop {t_1=0}} \right.$

$\lim_{a \to \infty} \theta = -\frac{3.7}{0.0714}e^{-0.0714a} + \frac{3.7}{0.0714}e^{-0.0714(0)}\\\\ \lim_{a \to \infty} \theta = \frac{3.7}{0.0714}(1) = 51.82 rad$

Convert to rev:

$51.82 rad * \frac{1rev}{2\pi rad} = \boxed{8.25 rev}$

8 0

3 years ago

A brick is thrown upward from the top of a building at an angle of 10° to the horizontal and with an initial speed of 16 m/s. If

Basile [38]

Answer:

38.47 m

Explanation:

To find the height of the building, we will use the following equation

$y_f=y_i+v_{iy}t+\frac{1}{2}at^2$

Where yf is the final height, yi is the initial height, viy is the initial vertical velocity, t is the time, and a is the acceleration due to gravity.

If the brick is in flight for 3.1 s, we can say that when t = 3.1s, yf = 0 m. So, replacing

viy = (16 m/s)sin(10) = 2.78 m/s

a = -9.8 m/s²

we get

$0=y_i+2.78(3.1)+\frac{1}{2}(-9.8)(3.1)^2$

Solving for yi

$\begin{gathered} 0=y_i-38.48 \\ y_i=38.48\text{ m} \end{gathered}$

Therefore, the height of the building is 38.48 m

6 0

1 year ago

Last night, Shirley worked on her accounting homework for one and one half hours. During that time, she completed 6 problems. Wh

hodyreva [135]

Answer:

velocity in problems per hour = 4 per hour

so correct option is b. 4 per hour

Explanation:

given data

worked on homework time = 1.5 hour

completed = 6 problems

to find out

What is the velocity in problems per hour

solution

we know that Shirley solve complete 6 accounting homework problem in 1.5 hour so her velocity in problems per hour will be as

velocity in problems per hour = $\frac{complete\ problem}{time\ taken}$ ..................1

put here value we will get

velocity in problems per hour = $\frac{6}{1.5}$

velocity in problems per hour = 4 per hour

so correct option is b. 4 per hour

3 0

3 years ago

Read 2 more answers

Which of the Moon’s positions will cause the highest or spring tides?

olya-2409 [2.1K]

The answer is B. A and C. Spring tides occur during a new moon and full moon

7 0

4 years ago

Read 2 more answers