Question

1. an airplane accelerates at a rate of 15 m/s². how much time does it take to increase its speed from 100.0 m/s to 160.0 m/s?

a) 4.0 s
b) 17 s
c) 0.058 s
d) 0.25 s
e) 1.0 s

1. which one of the following is an si base unit?

a) gram
b) slug
c) newton
d) centimeter
e) kilogram

1. the distance $d$ that a certain particle moves may be calculated from the expression $d = at^3$ where $a$ is a constant; and $t$ is the elapsed time. what must the dimensions of $a$ be?

a) l/t
b) l
c) t
d) l/t²
e) l/t³

1. an object moving along a straight line has a negative velocity vector and a positive acceleration vector. which is true?

a) the object is speeding up in the negative direction.
b) the object is slowing down in the negative direction.
c) the object is speeding up in the positive direction.
d) the object is slowing down in the positive direction.
e) the object is not moving.

1. a projectile is launched at an angle of 25.0° and an initial velocity of 60.0 m/s on level ground. find the range of this projectile.

a) 281 m
b) 310 m
c) 4.70 m
d) 46.0 m
e) 172 m

Explanation:

Problem 1

Step1: Recall acceleration formula

Acceleration $a = \frac{v_f - v_i}{t}$, rearrange to $t = \frac{v_f - v_i}{a}$

Step2: Substitute given values

$v_f=160.0\ \text{m/s}$, $v_i=100.0\ \text{m/s}$, $a=15\ \text{m/s}^2$
$t = \frac{160.0 - 100.0}{15} = \frac{60}{15} = 4.0\ \text{s}$

Problem 2

Step1: Identify SI base units

SI base units include kilogram (mass), meter (length), second (time), etc. Gram, slug, centimeter are non-base; newton is a derived unit.

Problem 3

Step1: Equate dimensions of both sides

$[d] = [A][t^3]$, $[d]=L$, $[t]=T$, so $L = [A]T^3$

Step2: Solve for $[A]$

$[A] = \frac{L}{T^3}$

Problem 4

Step1: Relate velocity and acceleration signs

Negative velocity = moving in negative direction; positive acceleration = acceleration in positive direction (opposite to motion), so object slows down in negative direction.

Problem 5

Step1: Use projectile range formula

$R = \frac{v_0^2 \sin(2\theta)}{g}$, where $v_0=60.0\ \text{m/s}$, $\theta=25.0^\circ$, $g=9.81\ \text{m/s}^2$

Step2: Calculate $2\theta$ and $\sin(2\theta)$

$2\theta=50.0^\circ$, $\sin(50.0^\circ)\approx0.7660$

Step3: Substitute values

$R = \frac{(60.0)^2 \times 0.7660}{9.81} = \frac{3600 \times 0.7660}{9.81} \approx \frac{2757.6}{9.81} \approx 281\ \text{m}$

Answer:

1. a) 4.0 s
2. e) kilogram
3. e) $L/T^3$
4. b) The object is slowing down in the negative direction.
5. a) 281 m