Question

A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 20.0 m above the river, whereas the opposite side is a mere 2.1 m above the river. The river itself is a raging torrent 61.0 m wide.

A) How fast should the car be traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side?

B) What is the speed of the car just before it lands safely on the other side?

Answer 1

A) The car should be traveling at 31.9 m/s.

B) The speed of the car just before it lands on the other side is 37.0 m/s.

Step-by-step explanation:

Hi there!

A) Please see the attached figure for a better description of the problem. When the car reaches the other side of the river, its position vector will be r1 in the figure. The components of this vector are r1x and r1y.

If we place the origin of the frame of reference at the edge of the cliff, the components of the vector r1 will be:

r1x = 61.0 m

r1y = -20.0 m + 2.1 m = -17.9 m

The equations for the x and y-components of the position vector of the car are the following:

x = x0 + v0 · t

y = y0 + 1/2 · g · t²

Where:

x = horizontal position at a time t.

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

y = vertical position at a time t.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

Using the equation of the y-component of r1, we can find the time it takes the car to reach the other side of the river. We have to find the time at which the vector r1y is -17.9 m:

y = y0 + 1/2 · g · t² (y0 = 0 because the origin of the frame of reference is located at the edge of the cliff).

y = 1/2 · g · t²

-17.9 m = -1/2 · 9.8 m/s² · t²

-17.9 m / -4.9 m/s² = t²

t = 1.91 s

Now, using the equation of the x-component, we can find the initial velocity. We know that at t = 1.91 s, the horizontal component of the vector r1 is 61.0 m:

x = x0 + v0 · t (x0 = 0 because the origin of the frame of reference is located at the edge of the cliff).

x = v0 · t

61.0 m = v0 · 1.91 s

v0 = 61.0 m / 1.91 s = 31.9 m/s

The car should be traveling at 31.9 m/s.

B) The equation of the velocity vector of the car is the following:

v = (v0, g · t)

The horizontal component of the velocity vector is v0, 31.9 m/s.

Let's calculate the value of the vertical component:

vy = g · t

vy = -9.8 m/s² · 1.91 s

vy = -18.7 m/s

Then, the velocity vector of the car just before it lands on the other side is the following:

v = (31.9, -18.7) m/s

The magnitude of this vector is calculated as follows:

|v| = √[(31.9 m/s)² + (-18.7 m/s)²]

|v| = 37.0 m/s

The speed of the car just before it lands on the other side is 37.0 m/s.