A 1000-kg car is driving toward the north along a straight road at a speed of 20.0 m/s. The driver applies the brakes and the car comes to a rest in a distance of 140 m. What is the constant force applied to the car to bring it to rest?
first:-
X=X1+0.5(Vx+V1x)t
140=0+0.5(20)t
t=14s
second:-
a=v/t=20/14=1.43m/s2
F=m*a=1000*1.43=1430N to the south
Well, applying the brakes is like telling the car, "Hey, slow down! Take a break!" But in order to bring the car to rest, a force opposite to the direction of the car's motion is needed. So let's calculate the force using the good ol' Newton's second law:
The initial velocity of the car is 20.0 m/s, and it comes to rest, so the final velocity is 0 m/s. The distance traveled while braking is 140 m.
First, we can find the time it takes for the car to stop using the equation:
vf = vi + at
0 m/s = 20.0 m/s + at
Solving for time, we get:
t = -20.0 m/s / a
Now, we can calculate the acceleration using the equation:
d = vi * t + 0.5 * a * t^2
Substituting the known values:
140 m = 20.0 m/s * (-20.0 m/s / a) + 0.5 * a * (-20.0 m/s / a)^2
Now, let's solve this equation and see if we can come to a conclusion.
To solve this problem, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by the acceleration of the object.
Given:
Mass of the car (m) = 1000 kg
Initial velocity of the car (v) = 20.0 m/s
Final velocity of the car (v_f) = 0 m/s
Distance traveled under braking (d) = 140 m
First, we need to find the acceleration (a) of the car using the following equation:
v^2 = u^2 + 2ad
where:
v = final velocity (0 m/s)
u = initial velocity (20.0 m/s)
a = acceleration
d = distance traveled under braking (140 m)
Let's rearrange the above equation to solve for a:
0 = (20.0 m/s)^2 + 2a(140 m)
400 = 2a(140 m)
400 = 280a
Now, divide both sides of the equation by 280:
a = 400/280
a = 1.43 m/s^2
We have found the acceleration of the car under braking.
Next, we can find the net force (F_net) exerted on the car using Newton's second law:
F_net = ma
where:
m = mass of the car (1000 kg)
a = acceleration (1.43 m/s^2)
Now, let's calculate the net force:
F_net = 1000 kg * 1.43 m/s^2
F_net = 1430 N
Therefore, the constant force applied to the car to bring it to rest is 1430 N.
To find the constant force applied to the car, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a). In this case, the car is initially traveling at a constant velocity and comes to rest, so its acceleration will be negative.
First, let's find the acceleration of the car. We can use the kinematic equation:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as the car comes to rest)
u = initial velocity (20.0 m/s)
a = acceleration
s = distance traveled (140 m)
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
Substituting the given values, we get:
a = (0^2 - 20.0^2) / (2 * 140)
Simplifying:
a = (0 - 400) / 280
a = -1.43 m/s^2
Now that we have the acceleration, we can find the force using Newton's second law:
F = m * a
Substituting the given mass (1000 kg) and calculated acceleration (-1.43 m/s^2), we get:
F = 1000 kg * -1.43 m/s^2
F = -1430 N
Since force is a vector quantity, the negative sign indicates that the force is acting in the opposite direction of motion, which in this case is the opposite of the car's initial direction (north). Therefore, the constant force applied to the car to bring it to rest is 1430 N directed south.
force= mass*acceleration
What is acceleration?
vf^2=Vi^2+2ad solve for a.