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Greeley [361]
3 years ago
9

IIIIIIIIII neeeeeed helppppp again?

Physics
1 answer:
Juliette [100K]3 years ago
7 0

Answer:

The Principle of Progression

(I searched it up since I never learned this)

Explanation:

The principle of progression states that a person should start slowly and increase exercise gradually. Since Mandy is just getting started on her exercise routine, she should begin with a few workouts over a large span of time, then work her way up so she can do more workouts in a shorter span of time.

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The traffic on the freeway is moving at a constant speed of 24 m/sm/s. What distance does the traffic travel while the car is mo
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Incomplete question as there is so much information is missing.The complete question is here

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

Answer:

Distance traveled=240 m

Explanation:

Given data

Initial velocity of car v₀=0 m/s

Final velocity of car vf=24 m/s

Distance traveled by car S=120 m

To find

Distance does the traffic travel

Solution

To find the distance first we need to find time, for time first we need acceleration

So

(V_{f})^{2}=(V_{o})^{2}+2aS\\  So\\a=\frac{(V_{f})^{2}-(V_{o})^{2} }{2S}\\ a=\frac{(24m/s)^{2}-(0m/s)^{2} }{2(120)}\\a=2.4 m/s^{2}

As we find acceleration.Now we need to find time

So

V_{f}=V_{i}+at\\t=\frac{V_{f}-V_{i}}{a}\\t=\frac{(24m/s)-(0m/s)}{(2.4m/s^{2} )}\\t=10s

Now for distance

So

Distance=velocity*time\\Distance=(24m/s)*(10s)\\Distance=240m

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