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?