Respuesta :
The car travels a distance of 50ft before stopping, at a rate of 40 ft/sec when the brakes are applied. the car deceleration at a constant rate of 16 ft/sec^2.
Given , Initial speed of the coz ( Vo ) = 40 ft/ sec
Deceleration of the car (dv/dt ) = - 16 Ft/ sec^2
Final speed of the car (Vx ) = 0 Ft / sec
Let the distance travelled by the can be x at
any time t. Let u be the velocity at any time t .
Now , deceleration means rate of decrease of velocity.
So ,
dv/dt = - 16 ft / sec^2
decreasing with
Negative sign means the velocity is decreasing with time .
dv/dt = dv/dx(dx/dt)
dv/dx(dx/dt) = -16
dx/dt = v, so
v dv/dx = -16
vdv = -16 dx
Integrating both sides under the limit 40 to o
0 to x for 'x '. This gives
[tex]\int\limits^0_ a {v} \, dx = \int\limits^0_x {-16} \, dx[/tex]
Here, a=40
x=800/16
x=50ft.
The car travels a distance of 50ft
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