Unit 2: Kinematics in 1-D Exam Preparation 1. A bike first accelerates from 0.0 m/s to 5.0m/s in4.5 s, then continues at this constant speed for another 4.5 s What is the total distance traveled by the bike? A car traveling at 20 m/s when the driver sees a child standing in the road. Hetakes 0.80 sto react, then steps on the brakes and sows at 7.0 m/s. How far does the car go before it stops? Answer the following questions a out the c? whose motion is graphed below: 40 30 d (m) 20 10 -10 -20 20 40 60 80 t (s) a When was the car 20 m west of the origin? b. where was the car at 50 s? c. The car suddenly reversed direction. When and where did that occur? A car starts 200 m west of the town square and moves with a constant velocity of 15 m/s toward the east Draw a graph that represents the motion of the car a Where will the car be 10 minutes later? b. When will the car reach the town square? Atthesame time theda in #4 left, atruck was 400 ment of the town squre moving west a aconstant velocity of 12 m/s 5. Add the trucksmoton to the graph you drew for question #4 b. Find the time where the car passed the truck. 6. A car is coasting backwards downhill at a speed of 3.0 m's when the driver gets the engine started. After 2.5 s the car is moving uphill a 4.5 m/'s Assuming that uphill is postive direction, what is the car's average acceleration? 7. A car slows from 22 m/s to 3.0 m/s at a constant rate of 2.1 m/s. How many seconds are required before the car is traveling 3.0 m's?
no ideas on any of these?
Probably too late for exam prep, then ...
1. To find the total distance traveled by the bike, we need to calculate the distance traveled during acceleration and the distance traveled at constant speed.
First, let's calculate the distance traveled during acceleration:
Initial velocity, u = 0.0 m/s
Final velocity, v = 5.0 m/s
Time taken, t = 4.5 s
Using the equation: distance = (initial velocity + final velocity) * time / 2
Distance during acceleration = (0 + 5.0) * 4.5 / 2 = 11.25 m
Next, let's calculate the distance traveled at constant speed:
Velocity, v = 5.0 m/s
Time taken, t = 4.5 s
Distance at constant speed = velocity * time = 5.0 * 4.5 = 22.5 m
Finally, the total distance traveled by the bike is the sum of the distances during acceleration and at constant speed:
Total distance = Distance during acceleration + Distance at constant speed = 11.25 + 22.5 = 33.75 m
Therefore, the total distance traveled by the bike is 33.75 m.
2. To find the distance the car goes before it stops, we need to calculate the distance traveled during the reaction time and the distance traveled while slowing down.
Initial velocity, u = 20 m/s
Reaction time, t_reaction = 0.80 s
Braking velocity, v_braking = 7.0 m/s
Using the equation: distance = (initial velocity + final velocity) * time / 2
Distance during reaction time = (20 + 0) * 0.80 / 2 = 8.0 m
Next, let's calculate the distance traveled while slowing down:
Velocity, v_final = 0 m/s
Acceleration, a = -2.1 m/s^2 (negative sign indicates deceleration)
Using the equation: v_final^2 = v_initial^2 + 2 * a * distance
0^2 = 7.0^2 + 2 * (-2.1) * distance
0 = 49 + (-4.2) * distance
-49 = -4.2 * distance
distance = -49 / -4.2 = 11.667 m (rounded to three decimal places)
Therefore, the car goes 11.667 m before it stops.
3a. The car was 20 m west of the origin at 10 s. (Using the graph)
3b. The car was at 40 m at 50 s. (Using the graph)
3c. The car reversed direction at approximately 60 s and was at 20 m east of the origin. (Using the graph)
4a. Since the car moves with a constant velocity of 15 m/s, it would be at the same position 10 minutes (600 seconds) later.
4b. The car started 200 m west of the town square and moves toward the east with a velocity of 15 m/s. So, it will reach the town square after t = distance / velocity = 200 / 15 = 13.33 s. Therefore, the car will reach the town square after 13.33 s.
5. To add the truck's motion to the graph, we need to plot its position on the x-axis and time on the y-axis. Since the truck starts 400 m west of the town square and moves at a constant velocity of 12 m/s, the graph would be a straight line starting at -400 m on the x-axis and having a positive slope.
To find the time where the car passed the truck, we can observe the graph. The point where the car's position becomes greater than the truck's position indicates the car passing the truck. By finding the respective time value, we can determine when it occurred.
6. To find the car's average acceleration, we need to calculate the change in velocity and divide it by the time taken.
Change in velocity = Final velocity - Initial velocity
= 4.5 m/s - (-3.0 m/s) (considering uphill as positive)
= 7.5 m/s
Time taken = 2.5 s
Using the equation: average acceleration = change in velocity / time taken
Average acceleration = 7.5 m/s / 2.5 s
= 3.0 m/s^2
Therefore, the car's average acceleration is 3.0 m/s^2.
7. To find the time required before the car is traveling at 3.0 m/s, we need to calculate the time taken to decelerate from 22 m/s to 3.0 m/s at a constant rate.
Initial velocity, u = 22 m/s
Final velocity, v = 3.0 m/s
Acceleration, a = -2.1 m/s^2 (negative sign indicates deceleration)
Using the equation: v = u + a * t
3.0 = 22 + (-2.1) * t
3.0 - 22 = (-2.1) * t
-19 = (-2.1) * t
t = 19 / 2.1
≈ 9.05 s
Therefore, it takes approximately 9.05 seconds before the car is traveling at 3.0 m/s.
1. To find the total distance traveled by the bike, we need to find the distance traveled during acceleration and during constant speed separately and then add them together.
During acceleration:
Initial velocity (u) = 0.0 m/s
Final velocity (v) = 5.0 m/s
Time taken (t) = 4.5 s
Using the equation: distance = (initial velocity + final velocity) / 2 * time
Distance during acceleration = (0.0 + 5.0) / 2 * 4.5 = 11.25 m
During constant speed:
Velocity (v) = 5.0 m/s
Time taken (t) = 4.5 s
Distance during constant speed = velocity * time = 5.0 * 4.5 = 22.5 m
Total distance traveled = distance during acceleration + distance during constant speed = 11.25 + 22.5 = 33.75 m
Therefore, the total distance traveled by the bike is 33.75 meters.
2. To find how far the car goes before it stops, we need to find the distance covered during the reaction time and the distance covered during deceleration.
Reaction time (t1) = 0.80 s
Initial velocity (u) = 20 m/s
Final velocity (v) = 0 m/s (since the car stops)
Acceleration (a) = (v - u) / t1 = (0 - 20) / 0.80 = -25 m/s^2 (negative because it's decelerating)
Using the equation: distance = (initial velocity + final velocity) / 2 * time
Distance during reaction time = (20 + 0) / 2 * 0.80 = 8.0 m
Using the equation: distance = (initial velocity^2 - final velocity^2) / (2 * acceleration)
Distance during deceleration = (20^2 - 7.0^2) / (2 * -25) = 53.9 m
Total distance = distance during reaction time + distance during deceleration = 8.0 + 53.9 = 61.9 m
Therefore, the car goes 61.9 meters before it stops.
3a. To determine when the car was 20 m west of the origin, we need to look at the position (d) on the graph. The position is negative when the car is west of the origin.
Looking at the graph, the car is 20 m west of the origin at around t = 2 s.
3b. To determine where the car was at 50 s, again, we look at the position (d) on the graph.
Looking at the graph, the car is at around d = 20 m at t = 50 s.
3c. To determine when and where the car suddenly reversed direction, we look for where the position (d) changes from positive to negative.
Looking at the graph, the car reverses direction at around t = 4.4 s and is located at around d = -10 m.
4a. The graph represents a car starting 200 m west of the town square and moving with a constant velocity of 15 m/s toward the east.
To represent this on a graph, we plot time (t) on the x-axis and position (d) on the y-axis. We start the graph at d = -200 m and draw a straight line with a slope of 15 m/s moving towards the positive d (east) direction.
4b. To determine where the car will be 10 minutes later, we convert the time to seconds (1 minute = 60 seconds).
10 minutes = 10 * 60 = 600 s
Since the car is moving at a constant velocity of 15 m/s toward the east, the position can be found using the equation:
Position (d) = initial position + (velocity * time)
Position (d) = -200 + (15 * 600) = 8800 m
Therefore, the car will be 8800 meters east of the starting position.
4c. To determine when the car will reach the town square, we need to find the time at which the car reaches the position of the town square (which is at d = 0).
Using the equation:
Position (d) = initial position + (velocity * time)
0 = -200 + (15 * time)
Solving for time:
15 * time = 200
time = 200 / 15
time ≈ 13.3 s
Therefore, the car will reach the town square at approximately 13.3 seconds.
5. To add the truck's motion to the graph from question #4, we need to plot the truck's motion on the same graph.
The truck starts 400 m west of the town square and moves west with a constant velocity of 12 m/s. Since the truck is moving west, its velocity will be negative.
We plot time (t) on the x-axis and position (d) on the y-axis. We start the graph at d = -400 m and draw a straight line with a slope of -12 m/s moving towards the negative d (west) direction.
To find the time when the car passes the truck, we look for the point where the position (d) of the car and the truck intersect on the graph.
6. The car's average acceleration can be found using the equation:
Acceleration (a) = (change in velocity) / (time)
Change in velocity = final velocity - initial velocity
Initial velocity (u) = -3.0 m/s (since it's moving backward)
Final velocity (v) = 4.5 m/s
Time (t) = 2.5 s
Acceleration (a) = (4.5 - (-3.0)) / 2.5
Acceleration (a) = 7.5 / 2.5
Acceleration (a) = 3.0 m/s^2
Therefore, the car's average acceleration is 3.0 m/s^2.
7. To find the time required for the car to travel at 3.0 m/s, we need to use the formula:
Final velocity (v) = initial velocity (u) + (acceleration * time)
Initial velocity (u) = 22 m/s
Final velocity (v) = 3.0 m/s
Acceleration (a) = -2.1 m/s^2 (negative since it's slowing down)
Rearranging the equation to solve for time (t):
time (t) = (v - u) / a
time (t) = (3.0 - 22) / -2.1
time (t) = -19.0 / -2.1
time (t) ≈ 9.05 s
Therefore, it takes approximately 9.05 seconds for the car to reach a velocity of 3.0 m/s.