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docker41 [41]
4 years ago
10

A commuter backs her car out of her garage with an acceleration of 1.40 m/s^2. (a) How long does it take her to reach a speed of

2.00 m/s? (b) If she then brakes to a stop in 0.800 s, what is her deceleration?
Physics
1 answer:
jonny [76]4 years ago
7 0

Answer:

a) It takes her 1.43 s to reach a speed of 2.00 m/s.

b) Her deceleration is - 2.50 m/s²

Explanation:

The equation of velocity for an object that moves in a straight line with constant acceleration is as follows:

v = v0 + a · t

Where:

v = velocty.

v0 = initial velocity.

a = acceleration.

t = time.

a) Using the equation of velocity, let´s consider that the car moves in the positive direction. Then:

v = v0 + a · t

2.00 m/s = 0 m/s + 1.40 m/s² · t

t = 2.00 m/s / 1.40 m/s²

t = 1.43 s

It takes her 1.43 s to reach a speed of 2.00 m/s

b) Let´s use again the equation of velocity, knowing that at t = 0.800 s the velocity is 0 m/s:

v = v0 + a · t

0 = 2.00 m/s + a · 0.800 s

-2.00 m/s / 0.800 s = a

a = -2.50 m/s²

Her deceleration is - 2.50 m/s²

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Speed of Dayshawn travelling towards his home is 12 m/s

Speed of her mom towards his school is 5 m/s

They both starts at same time so whenever they will meet on their path the sum of the distance covered by Dayshawn and distance covered by his mom must be equal to the total distance of school and home

Now let say they both meet after "t" time when they starts motion

so we can write the total distance between school and home as

d = v_1*t + v_2*t

here d = 6492 m

v_1 = 12 m/s = speed of dayshawn

v_2 = 5 m/s = speed of his mom

now by solving the above equation

6492 = 12t + 5 t

t = \frac{6492}{17}

t = 381.9 s

so they will meet after 381.9 s from start which will be 3.36 minutes from there start

Also at this time the distance covered by her mom will be

d_2 = v_2*t

d_2 = 5* 381.9 = 1909.4 m

so they will meet at distance 1909.4 m from their home

4 0
4 years ago
Object A moves with a constant velocity of -10 m/s and object B moves with a constant velocity of 5 m/s. Which object has the la
KonstantinChe [14]
Object A has the larger speed because speed is not a vector quantity, it is scalar. This means that directionality, or in this case the sign of the velocity, doesn’t matter.
Speed = |velocity|
Object a speed =|-10|=10 m/s
Object b speed =|5|=5 m/s

Object a is the answer.
8 0
3 years ago
You are helping your friend move a new refrigerator into his kitchen. You apply a horizontal force of 275 N in the positive x di
Volgvan

Answer:

f = 347.08 N

Explanation:

The frictional force exerted by the floor on the refrigerator is given as follows:

f = \mu R = \mu W

where,

f = frictional force = ?

μ = coefficient of static friction = 0.58

W = Weight of refrigerator = mg

m = mass of refrigerator = 61 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

f = \mu mg\\f = (0.58)(61\ kg)(9.81\ m/s^2)\\

<u>f = 347.08 N</u>

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3 years ago
The amount of force we need to move an object is proportional to the amount of work done - the __________ force used, the more w
Vanyuwa [196]

Answer:

stronger.

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7 0
3 years ago
The maximum speed of a mass m on an oscillating spring is vmax . what is the speed of the mass at the instant when the kinetic a
umka21 [38]
Let
A =  the amplitude of vibration
k =  the spring constant
m =  the mass of the object

The displacement at time, t, is of the form
x(t) = A cos(ωt)
where
ω =  the circular frequency.

The velocity is
v(t) = -ωA sin(ωt)

The maximum velocity occurs when the sin function is either 1 or -1.
Therefore
v_{max} = \omega A
Therefore
v(t) = -V_{max} sin(\omega t)

The KE (kinetic energy) is given by
KE = \frac{m}{2}v^{2} = \frac{m}{2} V_{max}^{2} sin^{2} (\omega t)

The PE (potential energy) is given by
PE = \frac{k}{2} x^{2} = \frac{k}{2} A^{2} cos^{2} (\omega t)

When the KE and PE are equal, then
v^{2} = \frac{k}{m} A^{2} cos^{2} (\omega t)

For the oscillating spring,
\omega ^{2} =  \frac{k}{m} \\ V_{max} = \omega A =  \sqrt{ \frac{k}{m} } A
Therefore
v^{2} =  \frac{k}{m}  \frac{m}{k} V_{max}^{2} cos^{2} ( \sqrt{ \frac{k}{m} t} ) \\ v = V_{max}  \,cos( \sqrt{ \frac{k}{m} t} )

Answer: v(t) = V_{max} cos( \sqrt{ \frac{k}{m} t} )

3 0
4 years ago
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