A 2319kg car is moving down a road with a slope grade of 11% and slowing down at a rate of 3.8m/s^2.Find the direction and magnitude of the frictional force ( define positive in the forward direction ie down the slope)
So the equation I have is f= ma+ mgsinθ. For my angle I get...
So basically given this problem:
a 1892kg car is moving down a road with slope (grade) of 12% while slowing down at a rate of 3.7m/s^2 .What is the direction and magnitude of the frictional force?(define positive in the forward direction)
Would my force be the tangent of the angle 12...
Sorry these were the numbers, I accidently mixed two probs together:/
a 1892kg car is moving down a road with slope (grade) of 12% while slowing down at a rate of 3.7m/s^2 .What is the direction and magnitude of the frictional force?(define positive in the forward direction)
Using...
a 1000 kg car is moving down a road with slope (grade) of 15% while slowing down at a rate of 3.7m/s^2 .What is the direction and magnitude of the frictional force?(define positive in the forward direction)
Using f=ma
I have the mass 1000 kg, the acceleration 3.7m/s^2, and slope 15%...
Zero? The frictional force would be zero. So, then the grade (the incline) was given just to make us think and try to picture it, I guess.
Ok, how about if we're given the grade and told that the the car was speeding up at 3m/s^2
I know the mass (1000kg) and I think theta would be the...
fictional force problem....given a slope(grade)???
A 1000kg car is moving down a road with a slope(grade) of 15% at a constant speed of 15m/s. What is the direction and magnitude of the frictional force?
So...
V= 15m/s
a= 0 (constant)
The slope really throws me off..I don't know...
Yes this is what I had, but I mentioned the #'s changed when i entered my answer so 7.25 became 5.16. It is late, thank you for your help... I wouldn't have gotten this far so thank u and good night :)
Ok, the #'s change each time but same question so, given y0=4ft(1.21m) v0y=5.16m/s v0x=2.41.
Quad eqtn 1.21+5.16t+(-4.91)t^2
End up with -5.16+-sqrt50.3/2(-4.9) T=1.25. So when I plug this into the x eqtn x=v0t. X=2.41•1.25.
X= 3.01m
I dont know I this is right, once I enter it...
Hmmm... I'm still getting it wrong. When I put into the quadratic it's 1.21m + 7.25 t + -4.9t2. Btw it's 1.21m from the 4ft given... I get a time that is a decimal so when I put it into the x eqtn it's very low.... I'm stuck
So the quadratic eqtn would give me the time it would take .... So I would then have to plug this into the eqtn x=x0+v0t+1/2at^2 but it would become x=x0+v0t since a for x equals zero... Right?
I'm still confused...if I try to solve for t what is my change in y? It says 4ft from the ground....but what about what it travels when it's is thrown up? Also wouldn't I have to make a= 9.81?
you throw a ball straight up in the air with a speed of 7.25m/s. The moment the ball leaves your hand you start running away at a speed of 2.59m/s. How far are you from the ball the moment it hits the ground.
So I've gotten better at these kinematics equations but this one has too many...
Can I use the eqtn v=v0+at instead with the initial velocity being zero, velocity being 100m/6.04s and t=6.04s? and solving for a?
and them plug it into the x-x0 eqtn?
In class we've used these: v=v0+at, x=x0+v0t+1/2at^2, and v^2=V0+2a(x-x0)
but from what I see I have an intitial velocity of 0, a distance and starting point which would be x0=0 and x=100m then I have two times..... I'm having trouble fitting this all in..
Sorry, my prof. has been going over how an object moving at twice the speed of that of another will travel 4 times as far (what I said before didnt make sense....but now that I look back at my notes it says that the "stopping" distance will be 4 times as far...which is a different concept...
*confused
A car speeding up from rest to constant accel. travels 100 meters in 6.04s. How far did it travel in the first 2.37 seconds?
So I figure if I divide distance by time 100/6.04 it will give me the velocity 16.5m/s. And since distance = vt I would do d=15.6*2.37 which equals 37.03m but...
lol, no she's walking but since the question says that the dog goes and comes back and continues to do so...I imagine that he covers a lot of ground...but yes, I guess if she is walking then he actually wouldnt make it back to the starting point at all.
I guess I'm...
So I would have 2.16=vt and 2.16=2vt but I dont have velocity for either one or time except for the initial time of 0...I don't feel any closer to solving the problem. We havent gone over anything like this in class so I'm very confused :(
Let me provide more information. The problem is:
a woman walks back to her campsite, 2.16 km away. Her dog runs ahead, at twice the speed, reaches the campground and turns around until it meets her again. Then, the dog proceeds to run back to the campground again. This continues until the...
I know that something with double the speed will travel 4 times as far. So does that mean you take the given distance and multiply it by 4? or take the distance and multiply it by 2 and square it? for example if its 3 meters would it be 3*4 or (2x3)^2
A car speeding up from rest to constant accel. travels 100 meters in 6.04s. How far did it travel in the first 2.37 seconds?
So I figure if I divide distance by time 100/6.04 it will give me the velocity 16.5m/s. And since distance = vt I would do d=15.6*2.37 which equals 37.03m but this is...
Need help with "motion" question..please!
Homework Statement
you are sitting in a car that is stopped. The moment a biker passes you, you start accelerating at 3 m/s2, while the bicycle continues to move along at a constant speed of 9 m/s. After what distance will you catch up with the...