A car moving initially at a speed of 80 km/h and weighing 13,000 N is brought to a stop ina distance of 61 m. Find(a) the braking force and (b) the time required to stop. Assuming the same braking force, find (c) the distance and (d) the time required to stop if the car were going 40 km/hr initially.
Vi = 80 km/ h * 1000/3600 = speed in m/s
find that
average speed during stop if deacceleration is constant = Vi/2
find that
time to stop = 61/(Vi/2)
find that - call it t
acceleration a = = - Vi/t
so
Force = m a = - m Vi/t
-----------------------------------
for Vi = (1/2) original Vi
then use F , same mass so same a
0 = new Vi + a t
that gives you new t
d = Vi t +(1/2) a t^2
or just use average velocity of Vi/2 and t
a)5368 N
b)5.5 s
c)15.24 m
d) 2.74 s
To find the braking force, we can use the equation:
Force = (mass) × (acceleration)
However, we need to find the acceleration first. To do this, we can use the equation:
final velocity^2 = initial velocity^2 + 2 × acceleration × distance
Given:
Initial velocity (u) = 80 km/h = 22.2 m/s
Final velocity (v) = 0 m/s
Distance (s) = 61 m
Using the equation, we can solve for acceleration:
0^2 = 22.2^2 + 2 × acceleration × 61
484.84 = 2 × acceleration × 61
acceleration = 484.84 / (2 × 61)
acceleration = 4 m/s²
(a) The braking force can be calculated using the equation:
Force = 13000 N
(b) The time required to stop can be found using the equation:
final velocity = initial velocity + acceleration × time
0 = 22.2 + 4 × time
-22.2 = 4 × time
time = -22.2 / 4
time = -5.55 s (Ignoring the negative sign)
(c) To find the distance required to stop if the car were initially going at 40 km/hr, we can use the equation:
initial velocity (u) = 40 km/h = 11.1 m/s
final velocity (v) = 0 m/s
acceleration = 4 m/s²
Using the same equation as before:
0^2 = 11.1^2 + 2 × 4 × distance
0 = 123.21 + 8 × distance
distance = -123.21 / 8
distance = -15.40 m (Ignoring the negative sign)
(d) The time required to stop can be found using the equation:
0 = 11.1 + 4 × time
time = -11.1 / 4
time = -2.78 s (Ignoring the negative sign)
To find the answers to the given questions, we need to apply the laws of motion, specifically Newton's second law of motion and the equations of motion.
(a) The braking force can be found using Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, since the car is brought to a stop, the net force is equal to the braking force. The equation is:
F = m * a
where F is the force, m is the mass, and a is the acceleration.
First, we need to find the mass of the car. We can use the equation:
weight = mass * acceleration due to gravity
mass = weight / acceleration due to gravity
The weight of the car is given as 13,000 N, and the acceleration due to gravity is approximately 9.8 m/s^2. Plugging in these values, we can calculate the mass:
mass = 13,000 N / 9.8 m/s^2 = 1326.53 kg (rounded to two decimal places)
Since the car is brought to a stop, the final velocity is 0 m/s. The initial velocity is given as 80 km/h, which needs to be converted to meters per second (m/s):
initial velocity = 80 km/h * (1000 m/1 km) * (1 h/3600 s) = 22.22 m/s (rounded to two decimal places)
Using the equation of motion:
final velocity^2 = initial velocity^2 + 2 * acceleration * distance
We can rearrange the equation to solve for acceleration:
acceleration = (final velocity^2 - initial velocity^2) / (2 * distance)
Plugging in the values, we get:
acceleration = (0^2 - 22.22^2) / (2 * 61 m) = -8.07 m/s^2 (rounded to two decimal places; the negative sign indicates deceleration)
Finally, we can calculate the braking force:
F = m * a = 1326.53 kg * -8.07 m/s^2 = -10,709.71 N (rounded to two decimal places)
Therefore, the braking force is approximately -10,709.71 N.
(b) To find the time required to stop, we can use the equation of motion:
final velocity = initial velocity + acceleration * time
Since the final velocity is 0 m/s, we can rearrange the equation to solve for time:
time = (final velocity - initial velocity) / acceleration
Plugging in the values, we get:
time = (0 m/s - 22.22 m/s) / -8.07 m/s^2 = 2.75 s (rounded to two decimal places)
Therefore, the time required to stop is approximately 2.75 seconds.
(c) To find the distance required to stop if the car were initially moving at 40 km/h, we can use the same calculations, but with different initial velocity.
First, we need to convert the initial velocity to m/s:
initial velocity = 40 km/h * (1000 m/1 km) * (1 h/3600 s) = 11.11 m/s (rounded to two decimal places)
Using the equation of motion, we can calculate the distance:
distance = (final velocity^2 - initial velocity^2) / (2 * acceleration)
Plugging in the values, we get:
distance = (0^2 - 11.11^2) / (2 * -8.07 m/s^2) = 6.86 m (rounded to two decimal places)
Therefore, if the car were initially moving at 40 km/h, it would require approximately 6.86 meters to stop.
(d) To find the time required to stop if the car were initially moving at 40 km/h, we can use the same calculations, but with different initial velocity.
Using the equation of motion, we can calculate the time:
time = (final velocity - initial velocity) / acceleration
Plugging in the values, we get:
time = (0 m/s - 11.11 m/s) / -8.07 m/s^2 = 1.37 s (rounded to two decimal places)
Therefore, if the car were initially moving at 40 km/h, it would require approximately 1.37 seconds to stop.