1) A physics book is moved once around the
perimeter of a table of dimensions 1 m by 3
m.
What is the distance traveled?
2) A window washer drops a brush from a scaf-
fold on a tall office building.
What is the speed of the falling brush after
2.7 s? (Neglect drag forces.) The acceleration
due to gravity is 9.8 m/s2 .
Answer in units of m/s
How far has it fallen in that time?
answer in units of m
3) With an average acceleration of −0.96 m/s2,
how long will it take a cyclist to bring a bicycle
with an initial speed of 14.8 m/s to a complete
stop?
Answer in units of s
4) A car traveling initially at +5.7 m/s acceler-
ates uniformly at the rate of +0.79 m/s2 for a
distance of 262 m.
What is its velocity at the end of the accel-
eration?
Answer in units of m/s
What is its velocity after it accelerates for 132
m?
Answer in units of m/s
What is its velocity after it accelerates for 65
m?
Answer in units of m/s
PLEASE give right answers + explain how you got it. Thanks!!
1) To find the distance traveled, we first need to determine the perimeter of the table. The perimeter of a rectangle can be calculated by summing the lengths of all four sides. In this case, the table has dimensions of 1m by 3m. Therefore, the perimeter is given by:
Perimeter = 2(length + width) = 2(1m + 3m) = 2(4m) = 8m
So, the distance traveled around the perimeter of the table is 8 meters.
2) The speed of a falling object can be calculated using the equation:
Speed = initial velocity + (acceleration due to gravity * time)
In this case, we are neglecting drag forces, so the only force acting on the brush is the acceleration due to gravity, which is -9.8 m/s^2 (taking downwards as the positive direction). The initial velocity is 0 m/s since the brush is dropped from rest. Plugging these values into the equation and solving for speed:
Speed = 0 m/s + (-9.8 m/s^2) * 2.7 s = -26.46 m/s
Since speed is magnitude-only (no direction), we take the absolute value to obtain 26.46 m/s. Therefore, the speed of the falling brush after 2.7s is 26.46 m/s.
To find the distance fallen, we can use the kinematic equation:
Distance = initial velocity * time + (1/2) * acceleration * time^2
Since the brush is dropped from rest, the initial velocity is 0. Plugging in the values:
Distance = 0 m/s * 2.7 s + (1/2) * (-9.8 m/s^2) * (2.7 s)^2 = -34.722 m
Taking the absolute value, the brush has fallen 34.722 meters in that time.
3) The equation relating initial velocity, final velocity, acceleration, and time is:
Final velocity = Initial velocity + (acceleration * time)
In this case, the initial velocity is 14.8 m/s and the final velocity is 0 m/s (since the cyclist comes to a complete stop). The acceleration is -0.96 m/s^2 (negative since it is deceleration). Plugging in the values and solving for time:
0 m/s = 14.8 m/s + (-0.96 m/s^2) * time
Rearranging the equation:
-14.8 m/s = -0.96 m/s^2 * time
time = -14.8 m/s / -0.96 m/s^2 ≈ 15.42 s
Therefore, it will take approximately 15.42 seconds for the cyclist to bring the bicycle to a complete stop.
4) The final velocity can be calculated using the equation:
Final velocity = Initial velocity + (acceleration * distance)
For the first question, we can find the velocity at the end of the acceleration for a total distance of 262m. The initial velocity is +5.7 m/s (taking the direction into account) and the acceleration is +0.79 m/s^2. Plugging in the values:
Final velocity = 5.7 m/s + (0.79 m/s^2) * 262 m ≈ 212.98 m/s
Therefore, the velocity at the end of the acceleration is approximately 212.98 m/s.
For the second question, we can find the velocity after accelerating for a total distance of 132m. The initial velocity is +5.7 m/s and the acceleration is +0.79 m/s^2. Plugging in the values:
Final velocity = 5.7 m/s + (0.79 m/s^2) * 132 m ≈ 106.28 m/s
Therefore, the velocity after accelerating for 132m is approximately 106.28 m/s.
For the third question, we can find the velocity after accelerating for 65m. The initial velocity is +5.7 m/s and the acceleration is +0.79 m/s^2. Plugging in the values:
Final velocity = 5.7 m/s + (0.79 m/s^2) * 65 m ≈ 56.35 m/s
Therefore, the velocity after accelerating for 65m is approximately 56.35 m/s.