CAN SOMEONE PLEASE HELP ME WITH MY PHYSICS QUESTIONS? I NEED CORRECT ANSWERS ONLY

A driver enters a one-lane tunnel at 34.4 m/s. The driver then observes a slow-moving van 154 m ahead travelling (in the same di

navik [9.2K]

Answer:

Ans. B) 22 m/s (the closest to what I have which was 20.16 m/s)

Explanation:

Hi, well, first, we have to find the equations for both, the driver and the van. The first one is moving with constant acceleration (a=-2m/s^2) and the van has no acceletation. Let´s write down both formulas so we can solve this problem.

$X(van)=5.65t+154$

$X(driver)=34.4t+\frac{(-2)t^{2} }{2}$

or by rearanging the drivers equation.

$X(driver)=34.4t+t^{2}$

Now that we have this, let´s equal both equations so we can tell the moment in which both cars crashed.

$X(van)=X(driver)$

$5.65t+154=34.4t-t^{2}$

$0=t^{2} -(34.4-5.65)t+154$ $0=t^{2} -28.75t+154$

To solve this equation we use the following formulas

$t=\frac{-b +\sqrt{b^{2}-4ac } }{2a}$

Where a=1; b=-28.75; c=154

So we get:

$t=\frac{28.75 +\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=21.63s$ $t=\frac{28.75 -\sqrt{(-28.75)^{2}-4(1)(154) } }{2(1)}=7.12s$

At this point, both answers could seem possible, but let´s find the speed of the driver and see if one of them seems ilogic.

$V(driver)=V_{0} +at$ }

$V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(7.12s)=20.16\frac{m}{s}$ $V(driver)=34.4\frac{m}{s} -2\frac{m}{s^{2} } *(21.63s)=-8.86\frac{m}{s}$

This means that 21.63s will outcome into a negative speed, for that reason we will not use the value of 21.63s, we use 7.12s and if so, the speed of the driver when he/she hits the van is 20.16m/s, which is closer to answer A).

Best of luck

8 0

3 years ago

A 14,700 N car is traveling at 25 m/s. The brakes are applied suddenly, and the car slides to a stop. The average braking force

Vlada [557]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

vo = 25 m/sec
vf = 0 m/sec 
Fμ = 7100 N (Force due to friction) 
Fg = 14700 N 
With the force due to gravity, you can find the mass of the car: 
F = ma 
14700 N = m (9.8 m/sec²) 
m = 1500 kg 
Now, we can use the equation again to find the deacceleration due to friction: 
F = ma 
7100 N = (1500 kg) a 
a = 4.73333333333 m/sec² 
And now, we can use a velocity formula to find the distance traveled: 
vf² = vo² + 2a∆d 
0 = (25 m/sec)² + 2 (-4.73333333333 m/sec²) ∆d 
0 = 625 m²/sec² + (-9.466666666667 m/sec²) ∆d 
-625 m²/sec² = (-9.466666666667 m/sec²) ∆d 
∆d = 66.0211267605634 m 
∆d = 66.02 m

7 0

4 years ago

HELP ASAP!!! PLEASE!!!! A driver traveling on the highway at 100km/hr notices the speed limit changes to 50km/hr as a speed came

Talja [164]

Answer:

90m

Explanation:

Speed * time

100*0.9=90m

hope this helped xx

8 0

4 years ago

you are driving at 18m/s down Lyndale avenue. A car backs out a driveway 25 meters in front of you. You continue at that speed

zimovet [89]

Hello!

For this, first let's calculate time of stop:

t = (V - Vi) / a

Replacing:

t = (0 m/s - 18 m/s) / -5,4 m/s^2

Resolving:

t = -18 m/s / -5,4 m/s^2

t = 3,33 s + 0,25 s = 3,58 s

Now lets calculate distance traveled, with formula:

d = Vi*t + (a*t^2)/2

Replacing:

d = 18 m/s * 3,58 s + (-5,4 m/s^2 * (3,58 s)^2) /2

Resolving:

d = 64,44 m + (-34,604 m)

d = 29,83 m

Then, the vehicle will CRASH

7 0

3 years ago

Suppose you sound a 1056-hertz tuning fork at the same time you strike a note on the piano and hear 2 beats/second. You tighten

Sindrei [870]

Answer:

The frequency of the piano string is either 1053 HZ or 1059 HZ.

Explanation:

Here we know that frequency of beats is equal to the difference between the frequencies between two waves .

Given that frequency of tuning fork is 1056 HZ .

Let the frequency of the piano be ' f ' .

Given that number of beats = 3.

We know that | 1056 - f | = 3 ;

⇒ 1056- f = ±3,

Upon solving this we get

f = 1056-3 and 1056 + 3

⇒ f = 1053 or 1059 .

6 0

3 years ago