Solving Kinematics Problems in 2D: Get Help Now!

[Mesmer17]

ok so here are some problems that I need help with.. I will explain how I did them, and if anyone can tell me why I am getting them wrong and what I should do to fix it would be great.. Thanks...
1.)A wild horse starts from rest and runs in a straight line 29 degrees north of west. After 34s of running in this direction, the horse has a speecd of 17m=m/s..
a.) north and east are positive lines, fine the component of the horses acceleration pointing along the north south line.

Here is what I did.. I found the average acceleration to be .5m/s^2.. then using the value .5 and the time 34s.. I plugged it into the equation Vxf=Vxo+AxTf.. so it would be Vxf=0+.5(34) which is 17..
b) find the component of the horses acceleration that points along the east west line..
I used the same equation but For Vo I used 17.. and used the rest of the numbers at before.. and I got..34.

Q2) When chasing a hare along flat ground,a grayhound leaps into the air at a speed of 12.3m/s at an angle 27degrees above the horizontal.
A) what is the range of his leap?
To do this I found, T setting 5.6(found from 12.3(sin27)) added to 1/2(-9.80)T equal to zero. The time comes out to 1.1s, then I used this multipiled by 12.3cos27.. to find the range... I just don't know if I did this right...

Q3) Car drives straight off a cliff 55m High, the impact of the car is 140m from the base of the cliff... how fast was the car traveling when it went over the cliff..
To do this one,I found the time by Using the Equation T=the sq.root of 2H/g.. so The sq.root of 2(55)/9.80.. to equal 3.4. then using the time, I used the equation Vyf=-g(T) so -9.80(3.4) which equals -33.3



PLEASE HELP ASAP.. LET ME KNOW WHAT I AM DOING WRONG..

THANKS

 

[HallsofIvy]

1.)A wild horse starts from rest and runs in a straight line 29 degrees north of west. After 34s of running in this direction, the horse has a speecd of 17m=m/s..
a.) north and east are positive lines, fine the component of the horses acceleration pointing along the north south line.

Here is what I did.. I found the average acceleration to be .5m/s^2.. then using the value .5 and the time 34s.. I plugged it into the equation Vxf=Vxo+AxTf.. so it would be Vxf=0+.5(34) which is 17..
b) find the component of the horses acceleration that points along the east west line..
I used the same equation but For Vo I used 17.. and used the rest of the numbers at before.. and I got..34.

In problem 1 you are asked to find the component of acceleration.
The formula you are using "Vxf=Vxo+AxTf" is for speed.
Since you know that the average acceleration is 0.5 m/s²,
at 29 degrees north of west, the north component is -0.5 cos(29) and the west component is -0.5 sin(29).

Q2) When chasing a hare along flat ground,a grayhound leaps into the air at a speed of 12.3m/s at an angle 27degrees above the horizontal.
A) what is the range of his leap?
To do this I found, T setting 5.6(found from 12.3(sin27)) added to 1/2(-9.80)T equal to zero. The time comes out to 1.1s, then I used this multipiled by 12.3cos27.. to find the range... I just don't know if I did this right...

The y component of position is 5.6T- (1/2)(9.8)T²= 0 which has two solutions: T= 0 (of course) and T= (2*5.6)/9.8= 1.1 seconds. The x coordinate is (12.3 cos(27)T= 12.5 meters. You copied the formula wrong here ("T" instead of "T²) but that looks like what you are doing.

Q3) Car drives straight off a cliff 55m High, the impact of the car is 140m from the base of the cliff... how fast was the car traveling when it went over the cliff..
To do this one,I found the time by Using the Equation T=the sq.root of 2H/g.. so The sq.root of 2(55)/9.80.. to equal 3.4. then using the time, I used the equation Vyf=-g(T) so -9.80(3.4) which equals -33.3

Assuming the car drove horizontally off the cliff at speed v0, it's horizontal position will vo T and its vertical position will be
55- (1/2)(9.8)T². It hits when 55- (1/2)(9.8)T²= 0 or T= [sqrt](2*55/9.8), atain what you have.
However, "Vyf= -g(T)" appears to be the vertical speed of the car as it hit the base of the cliff. You were asked for the speed of the car "when it went over the cliff"- its initial speed.
Since you now know T= 3.4 seconds and vo T= 140, you can solve for
v0= 140/T= 140/3.4= 41.4 m/s.

 

[M98Ranger]

It looks like you have a good understanding of the equations and concepts involved in these problems. However, there are a few mistakes that may be causing your incorrect answers.

In problem 1a, you found the average acceleration to be 0.5 m/s^2, but this is actually the acceleration in the west direction. The question asks for the component of acceleration along the north-south line, which would be 0 m/s^2 since the horse is not accelerating in that direction.

In problem 1b, you used the same equation but incorrectly used the final velocity as the initial velocity. The correct equation to use would be Vxf = Vxo + Axt, where Vxo is the initial velocity in the east direction (which is 0 m/s) and Axt is the acceleration in the east direction. Therefore, the correct answer would be 0 m/s^2.

In problem 2, you used the correct equations, but you made a calculation error when finding the time. The correct equation to use would be T = 2Voy/g, where Voy is the vertical component of the initial velocity (12.3 m/s * sin27). This would give you a time of 1.4 seconds, and using this time in your range calculation would give you a correct answer of approximately 16.8 m.

In problem 3, you correctly used the equation for time, but you made a mistake when finding the final vertical velocity. The correct equation to use would be Vyf = Voy - gt, where Voy is the initial vertical velocity (which is 0 m/s). This would give you a final vertical velocity of -33.2 m/s, which is close to your answer but not exactly the same.

Overall, it seems like you have a good grasp on the concepts and equations involved in these problems. Just be sure to double check your calculations and make sure you are using the correct values in each equation. It may also be helpful to draw diagrams to visualize the problem and help you keep track of the different components and directions involved. Good luck!