Respuesta :
Answer:
a) The initial velocity should be 24.0 m/s
b) The final speed is 31.0 m/s
Explanation:
The equations for the position and velocity of the car are as follows:
r = (x0 + v0x · t, y0 + v0y · t +1/2 · g · t²)
v = (v0x, v0y + g · t)
Where:
r = vector position of the car
v0x = initial horizontal velocity
v0y = initial vertical velocity
t = time
y0 = initial vertical position
g = acceleration de to gravity
v = velocity at time t
Please, see the figure for a better understanding of the problem. Notice that the origin of the frame of reference is located at the edge of the bridge so that y0 = 0 and x0 = 0.
We know the components of the position vector (r final in the figure) when the car reaches the other side of the river:
r final = (48.0 m, -19.5 m)
Then, using the equations for the x and y-components of the vector r final, we can obtain the initial velocity.
First, let´s find the time it takes the car to reach the other side:
y = y0 + v0y · t +1/2 · g · t² (y0 and v0y = 0)
y = 1/2 · g · t²
-19.5 m = -1/2 · 9.80 m/s² · t²
-19.5 m · -2 / 9.80 m/s² = t²
t = 2.00 s
Now, we can calculate the initial velocity because we know that at t = 2.00 s the x-component of the vector r final is 48.0 m.
x = x0 + v0x · t (x0 = 0)
48.0 m = v0x · 2.00 s
v0x = 24.0 m/s
b) When the car lands on the other side, its velocity is given by the velocity vector:
v = (v0x, v0y + g · t)
We already know the x-component of v, so let´s find the y-component:
vy = v0y + g · t (v0y = 0)
vy = -9.80 m/s² · 2.00 s = -19.6 m/s
The final speed will be the magnitude of the vector v:
v² = ( 24.0 m/s)² + (-19.6 m/s)²
v = 31.0 m/s
(a)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.
What is speed?
Speed is defined as the rate of change of the distance attained. it is a time-based quantity.
it is denoted by u for the initial speed while v for the final speed. its SI unit is m/sec.
The equation for the position vector will be
[tex]\rm \vec r = (x_0 + v_0xt, y_0 + v_0yt +\frac{1}{2} gt^2)[/tex]
r is the vector position of the car
vₓ is the initial horizontal velocity
[tex]\rm v_y[/tex] is initial vertical velocity
t is time
y₀ is the initial vertical position
g is the acceleration due to gravity
v is 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
[tex]\rm \vec r_{final} = (48.0 m, -19.5 m)[/tex]
Then, employing the equations for the x and y-components of the vector r final, we will obtain;
[tex]\rm y = y_0 + v_0yt +\frac{1}{2} gt^2 \\\\ \rm y =\frac{1}{2} gt^2\\\\[/tex]
[tex]\rm -19.5 m = -\frac{1}{2}\times 9.80t^2\\\\ \rm t^2= 4\\\\ \rm t = 2.00 s[/tex]
By estimating the initial velocity at t = 2.00
The value is x-component of the vector r final is 48.0 m.
[tex]\rm x = x_0 + v_0xt\\\\ \rm( x_0 = 0)\\\\ 48.0 = v_0x\times 2.00 s\\\\ \rm v_0x = 24.0 m/s[/tex]
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.
[tex]v = (v_0x, v_0y + gt)[/tex]
The y-component of v is given by;
[tex]\rm v_y = v_0y + gt \\\\ \rm (v_0y = 0)\\\\ \rm v_y = -9.80 \times 2.00 \\\\ \rm \ v_y = -19.6 m/s[/tex]
The resultant speed of the vector v will be :
[tex]\rm v^2 = ( 24.0 m/s)^2 + (-19.6 m/s)^2\\\\ \rm v= \sqrt{ ( 24.0 m/s)^2 + (-19.6 m/s)^2} \\\\ \rm v = 31.0\ m/s\\\\[/tex]
Hence the speeds of the car just before it lands on the other side will be 31.0 m/s.
To learn more about the speed refer to the link;
https://brainly.com/question/7359669
