Ask question

Ask question

All categories

1 year ago

6

As a runner crosses the finish line of a race, she starts decelerating from a velocity of 9 m/s at a rate of 2 m/s^2. Take the r

unner's velocity as she crosses the finish line to be in the positive direction. What is the runner’s displacement, in meters, during the first 5 seconds after crossing the finish line?
What is her velocity, in meters per second, 5 seconds after crossing the finish line?

1 answer:

Ksivusya [100]1 year ago

4 0

Answer:

- 1 m/s, 20 m

Explanation:

u = 9 m/s, a = - 2 m/s^2, t = 5 sec

Let s be the displacement and v be the velocity after 5 seconds

Use first equation of motion.

v = u + a t

v = 9 - 2 x 5 = 9 - 10 = - 1 m/s

Use second equation of motion

s = u t + 1/2 a t^2

s = 9 x 5 - 1/2 x 2 x 5 x 5

s = 45 - 25 = 20 m

You might be interested in

What charge accumulates on the plates of a 2.0-μF air-filled capacitor when it is charged until the potential difference across

enot [183]

Answer:

0.0002 C.

Explanation:

Charge: This can be defined as the ratio of current to time flowing in a circuit. The S.I unit of charge is Coulombs (C)

Mathematically, charge can be expressed as

Q = CV ................................. Equation 1

Where Q = amount of charge, C = capacitance of the capacitor, V = potential difference across the plates.

Given: C = 2.0-μF = 2×10⁻⁶ F, V = 100 V.

Substitute into equation 1

Q = 2×10⁻⁶× 100

Q = 2×10⁻⁴ C

Q = 0.0002 C.

The amount of charge accumulated = 0.0002 C

7 0

1 year ago

Read 2 more answers

To a stationary observer, a man jogs east at 2.5 m/s and a woman jogs west at 1.5 m/s. from the woman's frame of reference, what

neonofarm [45]

T o a stationary observer, a man jogs east at 2.5 m/s and a woman jogs west at 1.5 m/s. from the woman's frame of reference, what is the man's velocity? it is 4m/s east

5 0

2 years ago

Read 2 more answers

A spaceship is travelling at 20,000.0 m/s. After 5.0 seconds, the rocket thrusters are turned on. At the 55.0 second mark, the s

tankabanditka [31]

Answer:

$80 m/s^2$

Explanation:

The acceleration of an object is given by:

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval it takes for the velocity to change from u to v

For the rocket in this problem,

u = 20,000 m/s

v = 24,000 m/s

t = 55.0 - 5.0 = 50.0 s

Substituting,

$a=\frac{24000-20000}{50}=80 m/s^2$

7 0

1 year ago

Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L

antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

$\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}$

Similarly, for rod 2

$\alpha_2 = \dfrac{\tau_2}{I_2}.$

Now, the moment of inertia for rod 1 is

$I_1 = \dfrac{1}{3}ML^2$ ,

and the torque acting on it is (about the center of mass)

$\tau_1 = Mg\dfrac{L}{2};$

therefore, the angular acceleration of rod 1 is

$\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},$

$\boxed{\alpha_1 = \dfrac{3g}{2L} }$

Now, for rod 2 the moment of inertia is

$I_2 = \dfrac{1}{3}(2M)(2L)^2$

$I_2 = \dfrac{8}{3} ML^2,$

and the torque acting is (about the center of mass)

$\tau _2 = (2M)g \dfrac{(2L)}{2}$

$\tau _2 = 2MgL;$

therefore, the angular acceleration $\alpha_2$ is

$\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.$

$\boxed{\alpha_2 = \dfrac{3g}{4L}}$

We see here that

$\dfrac{3g}{2L} > \dfrac{3g}{4L}$

therefore

$\boxed{\alpha_1 > \alpha_2.}$

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0

1 year ago

A basketball rolls off a deck that is 3.2 m from the pavement. The basketball lands 0.75 m from the edge of the deck. How fast w

astra-53 [7]

Answer:

d). The value of y should be -32m

Vx=0.92 m/s

Explanation:

Using equation of motion uniform to y motion

$x_{f}=x_{o}+v_{o}*t+\frac{1}{2}*a*t^{2}\\x_{o}=0m\\v_{o}=0\\x_{f}=\frac{1}{2}*a*t^{2}\\x_{f}=-3.2m \\a=g=-9.8\frac{m}{s^{2}} \\$

So to find t that is the same time for all the motion

$t^{2}=\frac{2*x_{f}}{a} \\t=\sqrt{\frac{2*x_{f}}{a}}=\sqrt{\frac{2*-3.2m}{-9.8\frac{m}{s^{2}}}}\\t=0.808s$

The value of Xf=-3.2m because the g is negative from the axis

Now in the axis 'x' to find Vx

$Vx=\frac{0.75m}{0.808S}\\ Vx=0.92 \frac{m}{s}$

5 0

2 years ago