Question

1. a ball is dropped from rest from a tower and strikes the ground 110 m below. approximately how many seconds does it take the ball to strike the ground after being dropped? neglect air resistance.

a) 2.50 s
b) 3.50 s
c) 4.74 s
d) 12.5 s
e) 16.0 s

1. the figure shows a graph of the position x of two cars, c and d, as a function of time t. according to this graph, which statements about these cars must be true?

a) the cars meet at t = 10 s.
b) the magnitude of the acceleration of car c is greater than the magnitude of the acceleration of car d.
c) at t = 10 s, both cars have the same velocity.
d) the magnitude of the acceleration of car c is less than the magnitude of the acceleration of car d.
e) car d has a greater velocity than car c.

1. a car is initially traveling at 50.0 km/h. the brakes are applied and the car stops over a distance of 35 m. what was the magnitude of the cars acceleration while it was braking?

a) 5.4 m/s²
b) 2.8 m/s²
c) 36 m/s²
d) 71 m/s²
e) 9.8 m/s²

1. a train with a constant speed of 16 m/s passes through a town. after leaving the town, the train accelerates at 0.33 m/s² until it reaches a speed of 35 m/s. how far did the train travel while it was accelerating?

a) 0.029 km
b) 0.53 km
c) 1.5 km
d) 2.3 km
e) 3.0 km

1. what must be your average speed in order to travel 350 km in 5.15 hr?

a) 67.0 km/hr
b) 0.147 km/hr
c) 66.0 km/hr
d) 14.7 km/hr
e) 68.0 km/hr

Explanation:

Problem 9

Step1: List known values

Initial velocity $v_0 = 0$ m/s, displacement $x = 110$ m, acceleration $a = g = 9.8$ m/s²

Step2: Use kinematic equation

$x = v_0 t + \frac{1}{2} a t^2$
Substitute $v_0=0$: $110 = 0 + \frac{1}{2} \times 9.8 \times t^2$

Step3: Solve for $t$

$t^2 = \frac{2 \times 110}{9.8} = \frac{220}{9.8} \approx 22.45$
$t = \sqrt{22.45} \approx 4.74$ s

• For position-time graphs, slope = velocity, curvature = acceleration.
• At $t=10$s, the graphs cross, so cars meet (a is true).
• Car C has a steeper increasing slope (greater velocity change, higher acceleration magnitude) than D (b is true, d false).
• At $t=10$s, slopes differ, so velocities differ (c false).
• At $t=10$s, C's slope is steeper, so C has greater velocity (e false).

Step1: Convert units

Initial velocity $v_0 = 50.0$ km/h = $50.0 \times \frac{1000}{3600} \approx 13.89$ m/s; final velocity $v=0$; distance $x=35$ m

Step2: Use kinematic equation

$v^2 = v_0^2 + 2 a x$

Step3: Solve for $a$

$0 = (13.89)^2 + 2 \times a \times 35$
$a = -\frac{(13.89)^2}{70} \approx -\frac{192.9}{70} \approx -2.76$ m/s², magnitude ≈ 2.8 m/s²

Answer:

c) 4.74 s

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Problem 10