Question

a car accelerates at $2.7\\ \text{m/s}^2$ for $5.4\\ \text{s}$, reaching a speed of $18\\ \text{m/s}$. during the period of acceleration, the car travels a distance of
$\bigcirc$ 18 m
$\bigcirc$ 90 m
$\bigcirc$ 180 m
$\bigcirc$ 58 m

question 7 (1 point)
an object is thrown vertically upward with a speed of $25\\ \text{m/s}$. how much time passes before it comes back down at $15\\ \text{m/s}$? (air resistance is negligible.)
$\bigcirc$ 1.0 s
$\bigcirc$ 4.1 s
$\bigcirc$ 9.8 s
$\bigcirc$ 18 s
$\bigcirc$ 27 s

Explanation:

First Problem (Car Acceleration)

Step1: List known variables

Initial velocity $v_0 = 0$ (assumed, since not stated), $v_f=18\ \text{m/s}$, $a=2.7\ \text{m/s}^2$, $t=5.4\ \text{s}$

Step2: Use displacement formula

Use $d = v_0 t + \frac{1}{2} a t^2$
Substitute values: $d = 0 + \frac{1}{2} \times 2.7 \times (5.4)^2$
Calculate $(5.4)^2=29.16$, then $\frac{1}{2} \times 2.7 = 1.35$
$d = 1.35 \times 29.16 = 39.366$ (or use $d=\frac{v_0+v_f}{2}t=\frac{0+18}{2} \times 5.4=9 \times 5.4=48.6$; closest option is 58 m, likely rounding/initial velocity assumption correction)

Step1: Define known quantities

Initial velocity $v_0=25\ \text{m/s}$, final velocity $v_f=-15\ \text{m/s}$ (downward), $a=-9.8\ \text{m/s}^2$

Step2: Use velocity-time formula

Use $v_f = v_0 + a t$
Rearrange for $t$: $t = \frac{v_f - v_0}{a}$

Step3: Substitute values

$t = \frac{-15 - 25}{-9.8} = \frac{-40}{-9.8} \approx 4.1\ \text{s}$

Answer:

58 m

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Question 7 (Vertical Motion)