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goldenfox [79]
3 years ago
5

A velocity vs time graph what is the acceleration of the object

Physics
2 answers:
tigry1 [53]3 years ago
7 0

Answer:

If the object is moving with an acceleration of +4 m/s/s (i.e., changing its velocity by 4 m/s per second), then the slope of the line will be +4 m/s/s.

Explanation:

Dmitry [639]3 years ago
6 0

Answer:

The graph is missing, but ill try to answer this anyway.

The acceleration is the rate of change of the velocity; this means that if you know the velocity as a function of time; v(t), you can obtain the acceleration a(t) by differentiate v(t) with respect to time; this is:

a(t) = dv(t)/dt

Them if you have a graph of velocity versus time, you can find the equation of v(t) and then find the acceleration.

For example, if your graph of velocity vs time is a linear relation, then the acceleration will be the slope of the linear relation.

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3 years ago
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At 2:00 p.m. a car's speedometer reads 30 mi/h. At 2:30 p.m. it reads 50 mi/h. Show that at some time between 2:00 and 2:30 the
Gelneren [198K]

Answer:

The acceleration is exactly 40 mi/h² as shown

Explanation:

Given;

initial velocity of the car, u = 30 mi/h

final velocity of the car, v = 50 mi/h

change in velocity, ΔV = v - u

                               ΔV = 50 mi/h - 30 mi/h = 20 mi/h

initial time, t₁ = 2:00 PM

final time, t₂ = 2:30 PM

Change in time, Δt = t₂ - t₁

                           Δt = 2: 30 - 2:00 = 30 mins  = 0.5 hour

Acceleration is given as change in velocity per change in time;

a = ΔV /  Δt

a = \frac{20 \ mi/h}{0.5 \ h}\\\\a = 40 \ mi/h^2 \ (shown)\\\\

Therefore, the acceleration is exactly 40 mi/h² as shown.

6 0
3 years ago
A cat dozes on a stationary merry-go-round, at a radius of 5.7 m from the center of the ride. Then the operator turns on the rid
tatyana61 [14]

Answer:

0.76

Explanation:

we are given:

radius (r) =5.7 m

speed (s) = 1 revolution in 5.5 seconds

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coefficient of friction (Uk) = ?

 we can get the minimum coefficient of friction from the equation below

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m x r x ω^{2} = Uk x m x g

r x ω^{2} = Uk x g

Uk = \frac{ r x ω^{2} }{g}

where ω (angular velocity) = \frac{2π}{time}

= \frac{2π }{5.5} = 1.14

Uk = \frac{ 5.7 x 1.14^{2} }{9.8} = 0.76

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4 years ago
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Answer:

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Explanation:

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What happens when two forces act in the same direction?
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