A car starts from rest and accelerates from the beginning of an inclined ramp at

Question

A car starts from rest and accelerates from the beginning of an inclined ramp at rate of 3.0 m/s^2. The ramp is 50 m long & height 20 m. The car leaves the ramp and then continues as a projectile until hits the ground.
What is the magnitude and direction of cars velocity when it leaves the ramp?
What is the max height reached by car?
What is horizantal distance from the point where the car leaves the ramp to where it hits the ground?
What are the magnitude and direction of the car's velocity when it hits the ground?

Explanation / Answer

the variables that you know. x=51m xo=0 Vx=23cos42 m/s t=? y=0 yo=0 Voy=? Vy=-23sin42 m/s Ay=-9.8m/s^2 t=? These variables represent: x is the horizontal distance traveled. xo is the initial horizontal position. y is the vertical displacement yo is the initial vertical position. Vx is the speed in the x direction, which is constant due to a lack of acceleration in the x direction. Voy is the initial speed in the y direction. We will be looking for this term. Vy is the final speed in the y direction. Ay is the acceleration in the y direction. It is equal to the acceleration due to gravitiy. t is the time of the jump. Now, the problem is very easy. We need two equations: For the x direction we will use the constant velocity equation: x-xo=Vxt. For the y direction we will use the equation that relates final velocity to initial velocity, time, and acceleration: Vy=Voy+Ayt. Solve for t in the x equation: 51m-0=(23cos42 m/s)t t=51/23cos42 t=2.98s. Now, plug t into the y equation and find Voy: -23sin42=Voy+(-9.8)(2.98) -23sin42+29.24=Voy Voy=13.9m/s. Finally, Voy and Vx are the components of the initial velocity. To find the magnitude, square each, add them together, then take the square root: squareroot((13.9m/s)^2+(17.1m/s)^2) =squareroot(484) =22.0m/s. After you have the magnitude, place the components tail to tail, then draw in the other sides of the rectangle. The resultant vector will go from the origin across the rectangle to the opposite corner. Then you have a right triangle. Solve for the angle made with the horizontal: tan(theta)=Voy/Vx theta=arctan(13.9/17.1) theta=39.0 degrees. Your initial velocity vector is therefore: Vo=22.0m/s at 39.0 degrees above the horizontal.

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