a bus moving at 20m/s slows at a rate of 4m/s each second. how far does it travel while braking. i found the time which was 5 seconds.
A car accelerates from rest at a rate of 5 m/s/s for a period of 4.2 seconds and then moves at a constant speed for 8 seconds more. how far did it travel.
a ball is launched at an angle of 30 degrees from the horizontal at a speed of 1.8 m/s/ what is the range? i found that vix=1.55 and viy=.90 then what do i do?
1. To find the distance traveled by the bus while braking, we can use the equation:
distance = initial velocity x time + 0.5 x acceleration x time^2
First, let's find the acceleration. Since the bus is slowing down, the acceleration is negative. The rate of slowing down is 4m/s each second, so the acceleration is -4 m/s².
Now, we can plug in the values into the equation:
distance = 20m/s x 5s + 0.5 x (-4 m/s²) x (5s)²
distance = 100m + (-0.5 x 4 m/s² x 25s²)
distance = 100m - 50m
distance = 50 meters
So, the bus travels 50 meters while braking.
2. To find the total distance traveled by the car, we need to find the distance traveled during acceleration and the distance traveled during constant speed separately.
First, let's find the distance traveled during acceleration:
distance = initial velocity x time + 0.5 x acceleration x time^2
Since the car starts from rest, the initial velocity is 0:
distance = 0.5 x 5 m/s² x (4.2s)²
distance = 0.5 x 5 m/s² x 17.64s²
distance = 44.1 meters
Now let's find the distance traveled at a constant speed. First, we need to find the final velocity of the car after acceleration using the formula:
final velocity = initial velocity + acceleration x time
final velocity = 0 + 5 m/s² x 4.2s
final velocity = 21 m/s
Now, we can find the distance traveled during constant speed:
distance = velocity x time
distance = 21 m/s x 8s
distance = 168 meters
So the total distance traveled is the sum of distances during acceleration and constant speed:
total distance = 44.1m + 168m
total distance = 212.1 meters
So, the car travels 212.1 meters in total.
3. To find the range, we first need to find the time of flight.
time of flight = 2 x (initial velocity y) / g
where g is the acceleration due to gravity, which is approximately 9.81 m/s².
time of flight = 2 x 0.90 m/s / 9.81 m/s²
time of flight ≈ 0.1837 seconds
Now, we can find the range using the horizontal velocity:
range = initial velocity x (time of flight)
range = 1.55 m/s x 0.1837s
range ≈ 0.2847 meters
So, the range of the ball is approximately 0.285 meters.
To find how far the bus travels while braking, we can use the equation of motion:
distance = initial velocity * time + (1/2) * acceleration * time^2
Given:
Initial velocity (u) = 20m/s
Acceleration (a) = -4m/s^2 (negative sign indicates deceleration)
Time (t) = 5s
Plugging in the values into the equation, we get:
distance = 20m/s * 5s + (1/2)(-4m/s^2)(5s)^2
distance = 100m - 50m
distance = 50m
So, the bus travels a distance of 50 meters while braking.
Now, let's find the distance traveled by the car:
Since the car accelerates at a rate of 5m/s^2 for 4.2 seconds and then moves at a constant speed for 8 more seconds, we can calculate the total distance using two separate calculations.
1. Distance covered during acceleration:
Using the equation of motion:
distance = (1/2) * acceleration * time^2
Acceleration (a) = 5m/s^2
Time (t) = 4.2s
Plugging in the values, we get:
distance_acc = (1/2) * 5m/s^2 * (4.2s)^2
distance_acc = 0.5 * 5m/s^2 * 17.64s^2
distance_acc = 44.1m
2. Distance covered at constant speed:
Since the car moves at a constant speed after the acceleration phase, the distance can be calculated using:
distance = speed * time
Speed = 5m/s (which is the final speed reached during acceleration)
Time (t) = 8s
Plugging in the values, we get:
distance_const = 5m/s * 8s
distance_const = 40m
Now, we can find the total distance traveled by adding the distance covered during acceleration and distance covered at constant speed:
total_distance = distance_acc + distance_const
total_distance = 44.1m + 40m
total_distance = 84.1m
Therefore, the car travels a total distance of 84.1 meters.
For the projectile motion of the ball launched at an angle of 30 degrees with a speed of 1.8 m/s, you have already found the horizontal (vix) and vertical (viy) components of the initial velocity.
To find the range (distance traveled horizontally), you can use the equation:
range = horizontal velocity * time
Since the initial vertical velocity (viy) will continue to affect the horizontal motion, you need to find the time (t) it takes for the ball to reach the ground. You can use the equation for vertical motion:
vertical displacement = initial velocity * time + (1/2) * acceleration * time^2
In this case, the initial vertical displacement is 0 because the ball is launched horizontally. The acceleration due to gravity (g) is approximately 9.8 m/s^2.
0 = viy * t + (1/2) * (-9.8 m/s^2) * t^2
Simplifying the equation, you get:
-4.9t^2 + 0.9t = 0
Solving for t, you find two possible solutions: t = 0 or t = 0.184s. Since time cannot be negative or zero, we take t = 0.184s as the time it takes for the ball to reach the ground.
Now, using the time calculated, you can find the range:
range = vix * t
range = 1.55m/s * 0.184s
range ≈ 0.28m
Therefore, the range of the ball is approximately 0.28 meters.
To find the distance the bus travels while braking, you need to use the equation of motion:
d = v0t + (1/2)at^2
where:
d = distance traveled
v0 = initial velocity
t = time
a = acceleration
In this case, the initial velocity (v0) of the bus is 20 m/s, and the acceleration (a) is -4 m/s^2 (negative because it's slowing down).
You found the time (t) to be 5 seconds. Plug these values into the equation:
d = (20 m/s)(5 s) + (1/2)(-4 m/s^2)(5 s)^2
Simplify the equation:
d = 100 m + (1/2)(-4 m/s^2)(25 s^2)
= 100 m - 50 m
The distance traveled by the bus while braking is:
d = 50 meters.
Now let's move on to the second question:
To find the distance the car traveled, we have two parts to consider: the period of acceleration and the constant speed part.
For the period of acceleration, you can use the equation:
d1 = (1/2)at^2
where:
d1 = distance traveled during acceleration
a = acceleration given (5 m/s^2)
t = time during acceleration (4.2 seconds)
d1 = (1/2)(5 m/s^2)(4.2 s)^2
= (1/2)(5 m/s^2)(17.64 s^2)
= 44.1 m
Now, for the constant speed part, the distance traveled can be found using the equation:
d2 = v*t
where:
d2 = distance traveled during constant speed
v = constant speed (which is the final velocity after acceleration)
t = time during constant speed (8 seconds)
Since the car is moving at a constant speed, the final velocity is the same as the velocity reached during acceleration. So, the final velocity (v) can be found using:
v = a*t
v = (5 m/s^2)(4.2 s)
= 21 m/s
Now, calculate the distance during constant speed:
d2 = (21 m/s)(8 s)
= 168 m
The total distance traveled by the car is:
d = d1 + d2
= 44.1 m + 168 m
= 212.1 meters.
Let's move on to the third question:
To find the range of the ball, we can use the projectile motion equations. The range is the horizontal distance traveled by the ball.
The horizontal component of the initial velocity (vix) is 1.55 m/s, and the vertical component of the initial velocity (viy) is 0.90 m/s.
The time of flight (T) can be calculated using the equation:
T = 2viy / g
where:
g = acceleration due to gravity (9.8 m/s^2)
T = (2 * 0.90 m/s) / 9.8 m/s^2
= 0.18 s
Now, calculate the range (R) using the equation:
R = vix * T
R = 1.55 m/s * 0.18 s
= 0.279 m
So, the range of the ball is approximately 0.279 meters.