Answer:
264.7 m
Explanation:
The net force felt by the car travelling along the slope is equal to the component of the weight parallel to the slope, so:
[tex]F=-mg sin \theta[/tex]
where
m is the mass of the car
g = 9.8 m/s^2 is the acceleration of gravity
[tex]\theta=10^{\circ}[/tex]
and the negative sign is due to the fact the force is opposite direction to the motion of the car
The acceleration of the car is therefore:
[tex]a=\frac{F}{m}=-gsin \theta=-9.8 sin 10^{\circ}=-1.7 m/s^2[/tex]
Now we can find how far the car went up the hill by using the equation:
[tex]v^2-u^2 = 2ad[/tex]
where
v = 0 is the final velocity of the car
u = 30 m/s is the initial velocity
g = -1.7 m/s^2 is the acceleration
d is the distance covered
Solving for d,
[tex]d=\frac{v^2-u^2}{2a}=\frac{0-30^2}{2(-1.7)}=264.7 m[/tex]
The distance up the hill that the car would coast before it starts to roll back down is 264.35 meters.
Given the following data:
Scientific data:
To calculate the distance up the hill that the car would coast before it starts to roll back down:
In accordance with Newton's Second Law of Motion, the net force acting on this car is equal to the vertical component of the weight that is parallel to the slope:
[tex]F=-mgsin \theta[/tex]
Note: The negative sign indicates that the net force is acting in opposite direction to the car's motion.
For the acceleration:
[tex]a = \frac{F}{m} = -gsin \theta\\\\a= -9.8 \times sin10\\\\a= -9.8 \times 0.1737\\\\a=-1.7023\;m/s^2[/tex]
For the distance:
We would apply the third equation of motion;
[tex]V^2 = U^2 + 2aS\\\\0^2 =30^2+2(-1.7023)S\\\\0=900-3.4046S\\\\S=\frac{900}{3.4046}[/tex]
S = 264.35 meters.
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