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A 200-kg car starts from restat position what (5 its speed at position C ? Elevations at posltions A. B and € are 4O-m, 15: mand 30-m respectively. (Frictionl...

Question

A 200-kg car starts from restat position what (5 its speed at position C ? Elevations at posltions A. B and € are 4O-m, 15: mand 30-m respectively. (Frictionless Path)

A 200-kg car starts from restat position what (5 its speed at position C ? Elevations at posltions A. B and € are 4O-m, 15: mand 30-m respectively. (Frictionless Path)



Answers

The car passes point $A$ with a speed of $25 \mathrm{m} / \mathrm{s}$ after which its speed is defined by $v=(25-0.15 s) \mathrm{m} / \mathrm{s}$ Determine the magnitude of the car's acceleration when it reaches point $B,$ where $s=51.5 \mathrm{m}$ and $x=50 \mathrm{m}$

In this problem, it has given that a car initially at best accelerate linearly at a constant rate for five seconds. Time is given five seconds in the final. The speed is after five seconds is 15 m for a second we have to find the acceleration of the car. So here we will use can domestic equation we Finally they call the initial plus acceleration in two times acceleration will be equal to the final minus. We initiated the worldwide time so this acceleration will be equal to 15 minus zero divided by time, which is five seconds. The exhibition will be called to three m per second square. Now we will see. Yeah. What was the acceleration of the car? So exploration of the car and is three m per second squared. This is the answer. Thank you.

Wait off the card is given and that it tickles too. The mass of the car, which is equals to 12.500 to the power of three kg And what I've done on eighties given with his 5 kg which is 5002 jewels. So basically, we have to come from the final speed in the first part. So we know that by the bar garage term work done Well, way change. It can technology that will be in a Socratic in the zero. So the federal energy we have half of the square. So what we've done is five 1000 is equal to one upon two into 2.5 into general, two out of three. So this and to be a square. So this one and this one get canceled and multiple ever too. So there's really on five. So this basically this one is going to get can seal. So be a square. He become nothing but 2 m parts again. And in the next part we have to get a great force on the on the excited on the car. So basically, the distance this one is given distance traveled in this time is even That is 25 m. So we know that b squared. You will do nothing. What you square plus to S and B zero in the U. S. Zero in this case would be a square you've got great is visit for that is about two times off a exploration into distance 25 this is and we can say that the acceleration comes out over there. This is was poor upon 15 m person going to square. So the force we have to learn in this will do nothing but mass times acceleration. That could be that will be 2.520 year off. Three times off, poured upon 50. So we can say this will become too over there. This will too. So we can say that to inter Tendons were off to Newton Force exerted on the car

So in this car, in this situation first, we need to draw the corporate taxes were say along the issues, the positive excesses. This is direction of the policy that's accident, This is east and the car which is initially moving along the west direction, but they speak he not X. And the acceleration of the car is the longer each direction. Let's see the X. So the velocity of the car at an instant lives the velocity of the cars that initiative velocity, that's the acceleration A. Into the time to guarantee. and the velocity of the car at D equals to 33 seconds is So the 37 seconds, the initial relatively is along the west direction. And we have they can easter's posture. Therefore, uh, we have to write -20 m/s. Plus the acceleration is along east, there for one m or second square with positive sign Into the time that is 37 seconds of the velocity of the car at equals to 37 seconds is 17 m possible. And the direction is along beast. The correct option is option B.

Per day. Now if why is equaling zero at point B? This means that why sub a is equaling 35.0 meters time sign of 50 degrees and this is giving us 26.8 meters now and then. Of course, why should be is equaling zero meters so we can see the potential energy at a with the equaling mg Weiss of A. And this is equaling 1.50 times 10 to the third. Or we can say 1500 kilograms multiplied by 9.80 meters per second squared multiplied by 26.8 meters. And this is giving us 3.94 times 10 to the fifth. Jules, the potential energy at B with the equaling mg. Why sabi? And of course, this would equal zero jewels. So essentially the change and the potential energy would be the potential energy at B minus the potential energy at a which is simply equaling negative 3.94 times 10 to the fifth Jules. So these would be our three answers for part four part B. Now we're going to say that why subsea is now equaling zero, which means that Weiss of A is gonna be equaling 50.0 meters sign of 50 degrees, giving us 38.3 meters. And then we have lice, a B equaling 15 teen 0.0 meters sign of 50 degrees and this is giving us 11.5 meters. So in this case we find that potential energy sub a would be equaling mg Weiss of A. So this would be 1500 kilograms multiplied by 9.80 meters per second squared multiplied by 38.3 meters. And this is giving us 5.63 times 10 to the fifth. Jules and we have the potential energy at B equaling mg Y c B. And this is giving us 1500 kilograms, 9.80 meters per second squared multiplied by 11.5 meters. And this is giving us 1.69 times 10 to the fifth Jules. And to find the change, we can say delta P. E. From a to be, this would simply be equaling. They got potential energy said be so. 1.69 minus 5.63 The potential energy at a and times 10 to the fifth Jules. And this is giving us negative 3.94 times 10 to the fort to the fifth. Jules Rather. So this would be our final answer. That is the end of the solution. Thank you for watching.


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