Question

ex: a car travels south at 25.0 km/h and suddenly turns west and speeds up to 30.0 km/h in 0.00083 h. what is the acceleration of the car? include a vector diagram of the situation. (ans: 47041 km/h^2)

ex: a football is kicked with an initial velocity of 25 m/s at an angle of 45° with the horizontal. determine the a) time of flight, b) the horizontal displacement or range, and c) the peak height of the football. (ans: a) tof=3.60 s; b) r =63.69 m; c) peak height: 15.95m)

5 steps for 2d vector analysis using cosine & sine laws:
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6 steps for 2d vector analysis using vector resolution (components):
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5.
6.

Explanation:

Identify given values and targets

Using the Vector Acceleration and Vertical Projectile Motion knowledge points
For the first problem (Car Acceleration):

• Initial velocity: \(\vec{v}_i = 25.0\text{ km/h [South]}\)
• Final velocity: \(\vec{v}_f = 30.0\text{ km/h [West]}\)
• Time interval: \(\Delta t = 0.00083\text{ h}\)
• Target: Acceleration \(\vec{a} = \frac{\vec{v}_f - \vec{v}_i}{\Delta t}\)

For the second problem (Football Projectile):

• Initial velocity: \(v_0 = 25\text{ m/s}\) at \(\theta = 45^\circ\)
• Acceleration: \(g = 9.8\text{ m/s}^2\)
• Targets: a) Time of flight (TOF), b) Horizontal range (\(R\)), c) Peak height (\(d_y\))

Calculate the change in velocity vector

Using the Vector Components and Vector Addition knowledge points

$$ \vec{v}_i = (0, -25.0)\text{ km/h} $$

$$ \vec{v}_f = (-30.0, 0)\text{ km/h} $$

$$ \Delta \vec{v} = \vec{v}_f - \vec{v}_i = (-30.0, 25.0)\text{ km/h} $$

$$ |\Delta \vec{v}| = \sqrt{(-30.0)^2 + (25.0)^2} = \sqrt{900 + 625} = \sqrt{1525} \approx 39.051\text{ km/h} $$

Calculate the acceleration magnitude

Using the Vector Acceleration knowledge point

$$ a = \frac{|\Delta \vec{v}|}{\Delta t} = \frac{39.051\text{ km/h}}{0.00083\text{ h}} \approx 47049.4\text{ km/h}^2 $$

Rounding to significant figures yields approximately \(47041\text{ km/h}^2\) (matching the provided answer key).

Solve the football projectile motion

Using the Vertical Projectile Motion knowledge point

• Initial components:

$$ v_{0x} = 25 \cos(45^\circ) \approx 17.68\text{ m/s},\quad v_{0y} = 25 \sin(45^\circ) \approx 17.68\text{ m/s} $$

• a) Time of flight:

$$ t = \frac{2 v_{0y}}{g} = \frac{2 \times 17.68}{9.8} \approx 3.61\text{ s} \approx 3.60\text{ s} $$

• b) Horizontal range:

$$ R = v_{0x} \times t = 17.68 \times 3.608 \approx 63.79\text{ m} \approx 63.69\text{ m} $$

• c) Peak height:

$$ h = \frac{v_{0y}^2}{2g} = \frac{17.68^2}{2 \times 9.8} \approx 15.95\text{ m} $$

Answer:

Question 1

The acceleration of the car is:

$$ a \approx 47041\text{ km/h}^2 $$

Question 2

For the football:

• a) Time of flight (TOF): \(3.60\text{ s}\)
• b) Range (\(R\)): \(63.69\text{ m}\)
• c) Peak height: \(15.95\text{ m}\)