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kirza4 [7]
3 years ago
11

These has two questions, any help would be much appreciated. Ignore the current answers.

Physics
1 answer:
Sati [7]3 years ago
4 0
Bro sorry I don’t know these answer but keeep trying
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A car and a motorcycle start from rest at the same time on a straight track, but the motorcycle is 25.0 m behind the car. The ca
sergeinik [125]

Answer:

t = 8.45 sec

car distance d = 132.09  m

bike distance d = 157.08 m

Explanation:

GIVEN :

motorcycle is 25 m behind the car , therefore distance need to covered by bike to overtake car is 25+ d, when car reache distance d at time t

for car

by equation of motion

d  = ut + \frac{1}{2}at^2

u = 0 starting from rest

d = \frac{1}{2}at^2

t^2 = \frac{2d}{a}

for bike

d+25 = 0 + \frac{1}{2}*4.40t^2

t^2= \frac{d+25}{2.20}

equating time of both

\frac{2d}{a} = \frac{d+25}{2.20}

solving for d we get

d = 132 m

therefore t is= \sqrt{\frac{2d}{a}}

t =  \sqrt{\frac{2*132}{3.70}}

t = 8.45 sec

each travelled in time 8.45 sec as

for car

d = \frac{1}{2}*3.70 *8.45^2

d = 132.09  m

fro bike

d = \frac{1}{2}*4.40 *8.45^2

d = 157.08 m

7 0
3 years ago
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
ExtremeBDS [4]

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

7 0
3 years ago
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