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rjkz [21]
3 years ago
6

Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.

Physics
2 answers:
Burka [1]3 years ago
7 0
Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.

At time  t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is  7/3 .              v1 = 7/3 m/s .

At time  t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
                                     v2 = zero .

At time  t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is  -3/5 .            v3 = -0.6 m/s .              
wel3 years ago
7 0
The slope or gradient of a line is a number that describes both the direction and the steepness of the line. For a distance-time graph, it signifies the velocity.

at t1=2.0s
v =  7/3 m/s 

At time  t2 = 4.0 s
v = zero

At time  t3 = 13 s
v = -0.6 m/s

Hope this answers the question. Have a nice day.
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A 4 kg toy car moves horizontally on a rough road with coefficient of kinetic friction 0.2. It accelerates from rest to 20 m/s i
attashe74 [19]

The total work done on the car is 784Joule.

<h3>What's the acceleration of the car?</h3>
  • As per Newton's equation of motion, V= U+at
  • U= initial velocity= 0 m/s

V= vinal velocity= 20m/s

t= time = 10s

a= acceleration

  • So, 20= 0+ 10a

=> a= 20/10= 2m/s²

<h3>What's the distance covered by the car in 10 seconds?</h3>
  • As per Newton's equation of motion,

V²-U² = 2aS

  • S= distance covered by the car
  • So, 20²-0=2×2×S=4S

=> 400= 4S

=> S= 400/4= 100m

<h3>What's the work done on the car due to frictional force?</h3>

Work done by frictional force= frictional force × distance

= (0.2×4×9.8)×100

= 784Joule

Thus, we can conclude that the work done on the car is 784Joule.

Learn more about the work done here:

brainly.com/question/25573309

#SPJ1

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