During a storm, a car traveling on a level horizontal road comes upon a bridge that has washed out. The driver must get to the other side, so he decides to try leaping the river with his car. The side of the road the car is on is 21.3 meters above the river, while the opposite side is only 1.8 meters above the river. The river itself is a raging torrent 48.0 meters wide.

Respuesta :

The speed of the car to land safely on the opposite side will be 24.0 m/s. The speed of the car just before it lands on the other side will be 31.0 m/s.

(a) Speed is defined as the rate of change of the distance attained. it is a time-based quantity.

The equation for the position vector will be;

= r = x₀ + v₀xt,y₀ + v₀yt + (1/2)gt²

where,

r = the vector position of the car

vₓ = the initial horizontal velocity

vₐ = initial vertical velocity

t = time

y₀ =the initial vertical position

g =the acceleration due to gravity

v = the velocity at time t

From the figure, the origin of the frame of reference is located at y₀ = 0 and x₀ = 0.

The given position vector from the figure;

= r(final) = (48 -19.5) m

Then, employing the equations for the x and y-components of the vector r final, we will obtain;

= y = y₀ + v₀yt + (1/2)gt²

= y = (1/2)gt²

= -19.5 m = (1/2) X 9.8 X t²

= t = 2 s

By estimating the initial velocity at t = 2.00

The value is x-component of the vector r final is 48.0 m.

= x = x₀ + v₀xt   [x₀ = 0]

= 48 = v₀x X 2

= v₀x  = 24 m/s

Hence the speed of the car to land safely on the opposite side will be 24.0 m/s.

(b) The speed of the car just before it lands on the other side will be 31.0 m/s.

= v = (v₀x,v₀y + gt)

The y-component of v is given by;

= vₐ = v₀y + gt   [v₀y = 0]

=  vₐ = -9.8 X 2

=  vₐ = -19.6 m/s

The resultant speed  of the vector v will be :

= v² = 24² + (-19.6)²

= v = √ (24² + (-19.6)²)

= v = 31 m/s

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The complete question:

During a storm, a car traveling on a level horizontal road comes upon a bridge that has washed out. The driver must get to the other side, so he decides to try leap- ing the river with his car. The side of the road the car is on is 21.3 m above the river, while the opposite side is only 1.8 m above the river. The river itself is a raging torrent 48.0 m wide. (a) How fast should the car be traveling at the time it leaves the road in order just to clear the river and land safely on the opposite side? (b) What is the speed of the car just before it lands on the other side?

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