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NeTakaya
2 years ago
9

How can you use the position-time graphs for two in-line skaters to determine if and when one in-line skater will pass the other

one?
Physics
1 answer:
Jet001 [13]2 years ago
8 0

Explanation:

Position-time graphs measure/express the position of a skater over time relative to the start or finish of the race (depends on how it is used). Note: are the skaters in line vertically or horizontally? Like is one directly behind the other or are they next to each other?

If the two skaters are in line horizontally with each other, then their position will be the same relative to the start or finish of the race. This means if one passes the other one, the position would be different for all times after they pass. On the graph, it would look like one single line at the start (as position is same) which splits into 2 (representing the new difference in position due to 1 passing the other.

If the two skaters are in line vertically, their lines on the graph will appear parallel to each other (assuming they are going same speed) because the position is changing at the same rate, one is just reaching the same point after the other. If the skater behind overtakes the one in front. The lines on the graph will cross and continue either in parallel but with the other line on top to represent the moment where their position is the same right before they pass and after, where the second skater is now in front.

Hope this helped!

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3 0
3 years ago
A 150kg motorcycle starts from rest and accelerates at a constant rate along with a distance of 350m. The applied force is 250N
mart [117]

Answer:

205N

Explanation:

The net force (F) is the difference between the applied force(F_{A}) and the kinetic frictional force(F_{R}). i.e

F = F_{A} - F_{R}    -----------------(i)

Note that;

F_{R} = μmg

Where;

μ = coefficient of kinetic friction

m = mass of the body

g = acceleration due to gravity = 10m/s²

Equation (i) then becomes;

F = F_{A} - μmg        -------------------(ii)

<em>Given from question;</em>

m = mass of motorcycle = 150kg

μ = 0.03

F_{A} = 250N

Substitute these values into equation (ii) as follows;

F = 250 - (0.03 x 150 x 10)

F = 250 - (45)

F = 205N

Therefore, the net force applied to the motorcycle is 205N

3 0
2 years ago
A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle
tatiyna

Answer:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

(a) find speed v

vertical displacement

-h= vsinA\times t-gt^2/2

putting values h=15 m, v=0.8

-15 = 0.8vt - 4.9t^2  ............. (1)

horizontal displacement d = vcosA×t = 0.6×v ×t

so v×t = d/0.6 = 40/0.6

plug it into (1) and get

-15 = 0.8\times40/0.6 - 4.9t^2

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) =  17.8 m/s

(b) If his speed was only half the value found in (a), where did he land?

v = 17.8/2 = 8.9 m/s

vertical displacement = -H =v sinA t - gt^2/2

⇒ 4.9t^2 - 8.9\times0.8t - 100 = 0

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m

3 0
2 years ago
What is the instantaneous speed of the object after the five seconds?
OverLord2011 [107]

Answer:

12.5 m/s

Explanation:

In a acceleration time graph the area under the curve gives the change in velocity of the object. Here object starts at rest and therefore initial velocity is 0. After 5 seconds acceleration is 5m/s2.

change in velocity=area under the curve

change in velocity= 0.5*acceleration* change in time

v-0=0.5*5*5

v=12.5 m/s

7 0
3 years ago
A projectile is fired at a velocity of 50 m/s straight into the air. What is the velocity of the projectile when the objects dis
Arisa [49]

Answer:

the final velocity of the object is 53.04 m/s.

Explanation:

Given;

initial velocity of the projectile, u = 50 m/s

displacement of the object, d = 16 m

let the final velocity of the object = v

Apply the following kinematic equation to determine the final velocity of the projectile.

v² = u² + 2gd

v² = 50² + (2 x 9.8 x 16)

v² = 2813.6

v = √2813.6

v = 53.04 m/s

Therefore, the final velocity of the object is 53.04 m/s.

8 0
3 years ago
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