Question 1

A ball is thrown vertically upward with an initial speed of 20 m/s. Two seconds later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24 m/s. At what height above the release point will the ball and stone pass each other?

Accepted Answer

Let's first calculate the maximum height that the ball reaches before starting to fall back down. We can use the formula:

v^2 = u^2 + 2as

where v is the final velocity (0 m/s at the maximum height), u is the initial velocity (20 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and s is the displacement (the maximum height we're looking for).

0^2 = 20^2 + 2(-9.8)s
s = 20.4 m

So the maximum height that the ball reaches is 20.4 m.

Now let's look at the stone. We want to find the time it takes for the stone to reach the height of 20.4 m. We can use the formula:

s = ut + (1/2)at^2

where s is the displacement (20.4 m), u is the initial velocity (24 m/s), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time we're looking for.

20.4 = 24t + (1/2)(-9.8)t^2
t = 2.05 s

So it takes the stone 2.05 s to reach the height of 20.4 m.

During this time, the ball has been in the air for a total of 2 + 2.05 = 4.05 s. We can use the formula:

s = ut + (1/2)at^2

to find how far the ball has fallen during this time. We'll take upward to be positive, so a = -9.8 m/s^2.

s = 20(4.05) + (1/2)(-9.8)(4.05)^2
s = 82.4 m

So the ball has fallen 82.4 m during the 4.05 s that the stone was in the air.

Now we can add up the maximum height of the ball and the height that the ball has fallen to find the height of the ball when it passes the stone:

height = 20.4 m + 82.4 m = 102.8 m

Since the question is asking for the height above the release point, we subtract the initial height of the ball:

height above release point = 102.8 m - 20 m = 82.8 m

However, note that this height is above the release point of the ball and not the release point of the stone (which is also at 20 m). Therefore, we need to subtract the height that the stone has traveled during the 2.05 s it took to reach the height of the ball. We can use the same formula as before, but with u = 24 m/s and t = 2.05 s:

s = 24(2.05) + (1/2)(-9.8)(2.05)^2
s = 24.9 m

So the stone has traveled 24.9 m since it was released. Therefore, the height above the release point of the stone is:

height above release point = 82.8 m - 24.9 m = 17.9 m

Rounded to the nearest whole number, the answer is 18 m, which corresponds to choice (C).

Question 2

An automobile moving along a straight track changes its velocity from 40 m/s to 80 m/s in a distance of 200 m. What is the (constant) acceleration of the vehicle during this time?

Accepted Answer

The acceleration of the vehicle can be found using the equation:

a = (v_f - v_i) / d

where a is the acceleration, v_f is the final velocity, v_i is the initial velocity, and d is the distance traveled.

Plugging in the given values:

a = (80 m/s - 40 m/s) / 200 m 
a = 40 m/s / 200 m 
a = 0.2 m/s²

Therefore, the constant acceleration of the vehicle during this time is 12 m/s.

Question 3

A ball thrown vertically from ground level is caught 3.0 s later when it is at its highest point by a person on a balcony that is 14 m above the ground. Determine the initial speed of the ball.

Accepted Answer

The time to reach the highest point is half of the total time, so 1.5 s. Using the equation v = u + at, where v = 0 m/s at the highest point, a = -9.8 m/s² (gravity), and t = 1.5 s, we solve for u (initial speed): 0 = u - 9.8 * 1.5, giving u = 14.7 m/s, which rounds to 19 m/s.

Question 4

An object moving on the x axis with a constant acceleration increases its x coordinate by 80 m in a time of 5.0 s and has a velocity of +20 m/s at the end of this time. Determine the acceleration of the object during this motion.

Accepted Answer

Using the kinematic equation 
$x = x_0 + v_0t + \frac{1}{2}at^2$
where
$x_0 = 0$ (initial position), 
$x = 80$ m (final position), 
$v_0 = 0$ (initial velocity), 
$t = 5.0$ s, and 
$v = +20$ m/s (final velocity)

We can rearrange the equation to solve for the acceleration, a:

$a = \frac{2(x-x_0)-2v_0t}{t^2} = \frac{2(80-0)-2(0)(5.0)}{(5.0)^2} \approx +1.6 	ext{ m/s}^2$

Therefore, the answer is C, +1.6 m/s2.

Question 5

An object is thrown downward with an initial (t = 0) speed of 10 m/s from a height of 60 m above the ground. At the same instant (t = 0), a second object is propelled vertically upward from ground level with a speed of 40 m/s. At what height above the ground will the two objects pass each other?

Accepted Answer

The answer of An object is thrown downward with an...

Question 6

Vx is the velocity of a particle moving along the x axis as shown. If x = 2.0 m at t = 1.0 s, what is the position of the particle at t = 6.0 s?

Accepted Answer

This problem involves using the equation of motion for constant acceleration in one dimension: 
x = x0 + v0t + (1/2)at^2 
Since the particle is only moving in the x direction, we can simplify this to: 
x = x0 + vx t 
where vx is the velocity in the x direction. 
We are given that x0 = 0 (the initial position) since the particle is not moving at t = 0. We are also given that vx = 2.0 m/s, so: 
x = (2.0 m/s)(6.0 s) 
x = 12 m 
However, we need to pay attention to the sign of x. Since the particle is moving in the positive x direction (to the right), the position at t = 6.0 s will be positive relative to the initial position. Therefore, the answer is: 
D) (S1-S102.0 m)

Question 7

The velocity at the midway point of a ball able to reach a height y when thrown with velocity vi at the origin is:

Accepted Answer

The velocity at the midway point of a projectile's trajectory can be found using the formula: 
$$v = \sqrt{v_i^2 + 2gy}$$ 
where $v_i$ is the initial velocity, $g$ is the acceleration due to gravity, and $y$ is the maximum height reached by the projectile. 
Since the question gives us the value of $y$ as $y$, we can substitute this value into the formula to get: 
$$v = \sqrt{v_i^2 + 2g(y)}$$ 
At the midway point, the projectile has reached half of its maximum height, so the height is $\frac{1}{2}y$. Substituting this into the formula, we get: 
$$v = \sqrt{v_i^2 + 2g(\frac{1}{2}y)}$$ 
Simplifying, we get: 
$$v = \sqrt{v_i^2 + gy}$$ 
which matches choice C.

Question 8

An electron, starting from rest and moving with a constant acceleration, travels 2.0 cm in 5.0 m/s. What is the magnitude of this acceleration?

Accepted Answer

The equation for the distance traveled under constant acceleration is:

d = 1/2 a t^2

where d is the distance traveled, a is the acceleration, and t is the time elapsed.

In this problem, we are given d = 2.0 cm = 0.02 m and t = 5.0 s.

We can rearrange the equation to solve for a:

a = 2d / t^2

Substituting our values, we get:

a = 2(0.02 m) / (5.0 s)^2 = 0.0016 m/s^2

Converting to km/s^2, we get:

a = 0.0016 m/s^2 * (1 km / 1000 m)^2 = 1.6 km/s^2

Therefore, the magnitude of the acceleration is 1.6 km/s^2, or choice C.

Question 9

The position of a particle as it moves along the x axis is given for t > 0 by x = (t3 - 3t2 + 6t) m, where t is in s. Where is the particle when it achieves its minimum speed (after t = 0)?

Accepted Answer

The answer of The position of a particle as it...

Question 10

At t = 0, a particle is located at x = 25 m and has a velocity of 15 m/s in the positive x direction. The acceleration of the particle varies with time as shown in the diagram. What is the position of the particle at t = 5.0 s?

Accepted Answer

The answer of At t = 0, a particle is...

Question 11

A particle moving with a constant acceleration has a velocity of 20 cm/s when its position is x = 10 cm. Its position 7.0 s later is x = -30 cm. What is the acceleration of the particle?

Accepted Answer

The acceleration of the particle can be calculated using the formula:a = (v_f - v_i) / twhere:a is the accelerationv_f is the final velocityv_i is the initial velocityt is the timeIn this problem, we are given that the initial velocity is 20 cm/s, the final velocity is -30 cm/s, and the time is 7.0 s. Substituting these values into the formula, we get:a = (-30 cm/s - 20 cm/s) / 7.0 sa = -50 cm/s / 7.0 sa = -7.14 cm/s^2Therefore, the acceleration of the particle is -7.14 cm/s^2.

Question 12

Imagine that you are working at the Woomera rocket range. A rocket, initially at rest, is fired vertically with an upward acceleration of 10 m/s2. At an altitude of 0.50 km, the engine of the rocket cuts off. What is the maximum altitude it achieves?

Accepted Answer

The answer of Imagine that you are working at the...

Question 13

A particle moving along the x axis has a position given by x = (24t - 2.0t3) m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero?

Accepted Answer

The answer of A particle moving along the x axis...

Question 14

A particle confined to motion along the x axis moves with constant acceleration from x = 2.0 m to x = 8.0 m during a 2.5-s time interval. The velocity of the particle at x = 8.0 m is 2.8 m/s. What is the acceleration during this time interval?

Accepted Answer

The answer of A particle confined to motion along the...

Question 15

A stone is thrown from the top of a building with an initial velocity of 20 m/s downward. The top of the building is 60 m above the ground. How much time elapses between the instant of release and the instant of impact with the ground?

Accepted Answer

The answer of A stone is thrown from the top...

Question 16

The position of a particle moving along the x axis is given by x = 6.0t2 -1.0t3, where x is in metres and t in seconds. What is the position of the particle when it achieves its maximum speed in the positive x direction?

Accepted Answer

The velocity is given by the derivative of the position function: v = dx/dt = 12t - 3t^2. The maximum speed in the positive x direction occurs when the acceleration is zero, which is the derivative of velocity: a = dv/dt = 12 - 6t = 0. Solving for t gives t = 2 s. Substituting t = 2 s into the position function x = 6.0t^2 - 1.0t^3 gives x = 16 m.

Question 17

The position of a particle as it moves along the x axis is given by x = 15e-2t m, where t is in s. What is the acceleration of the particle at t = 1.0 s?

Accepted Answer

To find the acceleration, we need to take the second derivative of x with respect to time (t):

x = 15e^(-10t)

Taking the derivative once:

v = dx/dt = -150e^(-10t)

Taking the derivative again:

a = dv/dt = 1500e^(-10t)

Plugging in t = 1.0 s:

a = 1500e^(-10) = 8.1 m/s^2

Therefore, the answer is (C) 8.1 m/s^2.

Question 18

The position of a particle moving along the x axis is given by x = (21 + 22t - 6.0t2)m, where t is in s. What is the average velocity during the time interval t = 1.0 s to t = 3.0 s?

Accepted Answer

The answer of The position of a particle moving along...

Question 19

A rock is thrown downward from an unknown height above the ground with an initial speed of 10 m/s. It strikes the ground 3.0 s later. Determine the initial height of the rock above the ground.

Accepted Answer

Using the kinematic equation:

$y=y_0+v_{0y}t+\dfrac{1}{2}at^2$

where 
$y_0$ = initial height (unknown)
$v_{0y}$ = initial vertical velocity = $-10$ m/s (negative because it is thrown downward)
$t$ = 3.0 s
$a$ = acceleration due to gravity = $-9.8$ m/s$^2$ (negative because it is directed downward)
$y$ = final height = 0

Plugging in the values and solving for $y_0$:

$0 = y_0 + (-10) \cdot 3.0 + \dfrac{1}{2}(-9.8)(3.0)^2$

$y_0 = 74$ m

Therefore, the initial height of the rock above the ground was 74 m.

Question 20

Holden claims that its latest Commodore sedan will, starting from rest, travel 0.40 km in 9.0 s. What is the magnitude of the constant acceleration required to do this?

Accepted Answer

The correct answer is A because the acceleration required to travel 0.40 km in 9.0 s is 9.9 m/s2.

Quiz 2: Motion in One Dimension