A car traveling 75 km/jam slows down at a constant 0,50 m/s2
just by “letting up on the gas.” Calculate (a) the distance the car
coasts before it stops, (b) the time it takes to stop, and (c) the
distance it travels during the first and fifth seconds.
Well, let's crunch some numbers!
(a) To calculate the distance the car coasts before it stops, we can use the equation:
distance = (initial velocity)^2 / (2 * acceleration)
Given that the initial velocity is 75 km/h and the deceleration is 0.50 m/s², we need to convert the initial velocity to meters per second:
75 km/h = 75,000 meters/3,600 seconds = 20.83 m/s
Plugging the values into the equation, we have:
distance = (20.83 m/s)^2 / (2 * 0.50 m/s²)
distance = 866.07 meters
So the car coasts for approximately 866.07 meters before coming to a stop.
(b) To find the time it takes for the car to stop, we use the equation:
final velocity = initial velocity + (acceleration * time)
Since the final velocity is 0, we can rearrange the equation to solve for time:
time = (final velocity - initial velocity) / acceleration
Plugging in the values:
time = (0 m/s - 20.83 m/s) / (-0.50 m/s²)
time ≈ 41.66 seconds
So it takes approximately 41.66 seconds for the car to stop.
(c) To calculate the distance the car travels during the first five seconds, we can use the equation:
distance = (initial velocity * time) + (0.5 * acceleration * time²)
For the first second:
distance = (20.83 m/s * 1s) + (0.5 * -0.50 m/s² * 1s²)
distance ≈ 20.83 meters
For the fifth second:
distance = (20.83 m/s * 5s) + (0.5 * -0.50 m/s² * 5s²)
distance ≈ 95.83 meters
So during the first second, the car travels approximately 20.83 meters, and during the fifth second, it covers around 95.83 meters.
Hope that answers your question with some mathematical comedy flair!
To calculate the distance the car coasts before it stops, we need to find the time it takes for the car to come to a stop. We can use the following equation of motion:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, since the car comes to a stop)
u = initial velocity (75 km/hour = 20.8 m/s)
a = acceleration (-0.50 m/s^2)
s = distance
Rearranging the equation, we get:
0 = (20.8)^2 + 2(-0.50)s
0 = 433.64 - s
Therefore, the distance the car coasts before it stops is s = 433.64 meters.
To calculate the time it takes to stop, we can use the formula:
v = u + at
Where:
v = final velocity (0 m/s)
u = initial velocity (20.8 m/s)
a = acceleration (-0.50 m/s^2)
t = time
Rearranging the equation, we get:
0 = 20.8 + (-0.50)t
t = -20.8 / -0.50
t = 41.6 seconds
Therefore, the time it takes for the car to stop is 41.6 seconds.
To calculate the distance the car travels during the first and fifth seconds, we can use the equation:
s = ut + (1/2)at^2
For the first second (t = 1 second):
s1 = (20.8)(1) + (1/2)(-0.50)(1^2)
s1 = 20.8 + (-0.25)
s1 = 20.55 meters
For the fifth second (t = 5 seconds):
s5 = (20.8)(5) + (1/2)(-0.50)(5^2)
s5 = 104 + (-6.25)
s5 = 97.75 meters
Therefore, the distance the car travels during the first second is 20.55 meters and during the fifth second is 97.75 meters.
To solve this problem, we need to use the equations of motion. There are four main equations of motion that relate distance, time, velocity, and acceleration:
1. v = u + at
2. s = ut + (1/2)at^2
3. v^2 = u^2 + 2as
4. s = vt - (1/2)at^2
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
s = distance
Given:
Initial velocity (u) = 75 km/h
Acceleration (a) = -0.50 m/s^2 (negative because the car is slowing down)
(a) Distance the car coasts before it stops:
In this case, the car is slowing down to a complete stop, so the final velocity (v) will be zero. We can use equation (1) to calculate the time it takes to come to a stop, and then use this time in equation (2) to calculate the distance.
Since the initial velocity (u) is given in km/h, we need to convert it to m/s:
u = 75 km/h * 1000 m/km * (1/3600) h/s
u = 20.83 m/s
Now we can find the time (t) to come to a stop:
Using equation (1): 0 = 20.83 m/s + (-0.50 m/s^2)t
t = -20.83 m/s / -0.50 m/s^2
t = 41.66 s
Now we can find the distance (s) using equation (2):
s = 20.83 m/s * 41.66 s + (1/2)(-0.50 m/s^2)(41.66 s)^2
s = 866.83 m (rounded to the nearest meter)
Therefore, the distance the car coasts before it stops is approximately 867 meters.
(b) Time it takes to stop:
As calculated above, it takes approximately 41.66 seconds for the car to come to a stop.
(c) Distance traveled during the first and fifth seconds:
To find the distance traveled during specific time intervals, we need to use equation (4).
For the first second (0-1 second):
Using equation (4): s = (0 m/s)(1 s) - (1/2)(0.50 m/s^2)(1 s)^2
s = 0 m
Therefore, the car does not travel any distance during the first second.
For the fifth second (0-5 seconds):
Using equation (4): s = (0 m/s)(5 s) - (1/2)(0.50 m/s^2)(5 s)^2
s = -6.25 m (negative because the car is slowing down)
Therefore, the car travels approximately 6.25 meters (backward) during the fifth second.
In summary:
(a) The distance the car coasts before it stops is approximately 867 meters.
(b) It takes approximately 41.66 seconds for the car to come to a stop.
(c) The car does not travel any distance during the first second, and it travels approximately 6.25 meters backward during the fifth second.
75000 meters/ 3600 seconds = 20.8 meters / second
a = -0.50 meters / second^2
v = Vi + a t
v = 20.8 - 0.50 t
stopped so v = 0
0.50 t = 20.8
t = 41.6 seconds (boring)
d = average speed * 41.6 seconds
= 10.4 * 41.6
= 433 meters (like forever )
during first second
Vi = 20.8
v at 1 s = 20.8 - 0.50 * 1 = 20.3
av v = 41.1/2 = 20.6 m/s
d = 20.6 * 1 = 20.6 meters
during fifth second
v at 4 seconds = 20.8 - 0.50 * 4 = 18.8 m/s
v at 5 seconds = 20.8 - 0.50 * 5 = 18.3 m/s
av v = 18.6 m/s
so in 1 seconds 18.6 meters