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In a stunt being filmed for a movie, a sports car overtakes a truck towing a ramp, drives up and off the ramp, soars into the air, and then lands on top of a flat trailer being towed by a second truck. The topsof the ramp and the flat trailer are the same height above the road, and the ramp is inclined 16 above the horizontal. Both trucks are driving at a constant speed of 11 m/s, and the flat trailer is 15 m from the end of the ramp. Neglect air resistance, and assume that the ramp changes the direction, but not the magnitude, of the car’s initial velocity. What is the minimum speed the car must have, relative to the road, as it starts up the ramp?

$v_{C R}=27.66 \mathrm{m} \cdot \mathrm{s}^{-1}$

Physics 101 Mechanics

Chapter 3

Kinematics in Two Dimensions

Motion in 2d or 3d

Rutgers, The State University of New Jersey

University of Washington

Simon Fraser University

University of Sheffield

Lectures

04:01

2D kinematics is the study of the movement of an object in two dimensions, usually in a Cartesian coordinate system. The study of the movement of an object in only one dimension is called 1D kinematics. The study of the movement of an object in three dimensions is called 3D kinematics.

10:12

A vector is a mathematical entity that has a magnitude (or length) and direction. The vector is represented by a line segment with a definite beginning, direction, and magnitude. Vectors are added by adding their respective components, and multiplied by a scalar (or a number) to scale the vector.

0:00

In a stunt being filmed fo…

13:27

. In a stunt being filmed …

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A movie stunt involves a c…

07:18

A stunt driver wants to ma…

01:52

A semitrailer is coasting …

05:41

(III) A stunt driver wants…

14:33

(1I) A stunt driver wants …

21:58

A truck with mass $m$ has …

03:04

In a movie, a stuntman pla…

So the question states that a ramp as well as some sort of flat top is moving to the right at 11 meters per second, and a car goes off the ramp toe land on top of the the D flat top. And we're trying to find how fast the car needs to go off the the ramp to land on the flat top that's 15 years away. So for this situation, we can pretend like the the ramp and the flat top aren't moving. So we're going to go into the frame where we're moving with the ramp as well as the laptop. So it's all stationary. So we know, um, how to figure out the, um, speed that the car must go if this whole situation is stationary. So to do this, we can separate our velocity vector into components. So, for example, here's our velocity V, which we're trying to find, which makes an angle of 60 degrees here. So we know this component here must be V. Times Co. Sign of 16 degrees and the vertical component must be V times sign of 16 degrees. And we know this because co sign of 16 degrees should be equal to the adjacent over iPod news and the adjacent in this case is the horizontal component. So we can solve for that and we will get this equation. Same thing goes for the vertical component. So now that we know the horizontal and vertical component, we can figure out how long the car is going to be in the air. Four. So to do this, we can use our Kitimat equations where state, um that the change in displacement in the Y direction is equal to the initial blast eight times the time plus one have times the acceleration times, the Times Square. So the displacement in the Y direction is going to be zero because it starts on the ground or the ramp and the flat top are in the same plane. So there's no upward or downward displacement, so this will be zero. The initial vertical displacement is going to be the our Sorry. The initial vertical velocity is going to be V sign of 16 degrees. We don't know what the time is. We'll just say t post 1/2 times the excel The acceleration, which is, uh, negative 9.8 meters per second squared times a Times Square so we can factor at a tea and we'll get zero is equal to be signed. 16 degrees plus one have times negative, 9.8 times t and then all this times t as well. So we'll see that tea can either be equal to zero or t can be equal to, um, negative to the sign of 16 degrees over the accelerated NIA acceleration, which is negative 9.8. So now that we know this the time that it takes for the ah car to reach the ramp, we can figure out the velocity as to travel at to, um, go at least 15 meters. So to do this, we know that the velocity in the X direction times sometime t should be equal to the displacement. So we know the velocity in the X direction is gonna be V times coastline of 16 degrees, So v co sign 16 degrees. We know the time because we calculated it up. Up above is to the sign 16 degrees over 9.8. And this is going to be, uh, equal to 15 meters. Now that we have this, we can combine our Weise. We get V squared times to co sign 16 degrees signed 16 degrees, divided by 9.8 is equal to 15 meters and so we can sell for V by multiplying by 9.8 almost sides and then dividing by two times coastline, 16 degrees time signed 16 years and then taking the square root. And that will give us a velocity of 16.655 meters per second. But this is from the point of view of where we're moving along with the, uh, the flat top in the ramp. So we need to add on the speed that this Ah, that the ramp in the flat topper moving at, which is 11 meters per second. And this will be the speed the car has to move relative to the ground. So the speed is gonna be equal to, um, 27 0.6 six theaters per second. And that's the final answer

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