Question

question 8 (1 point)
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
18 m
90 m
180 m
58 m
question 9 (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.)
1.0 s
4.1 s
9.8 s
18 s
27 s

Explanation:

Question 8

Step1: Find initial velocity

Use $v = u + at$, rearrange for $u$:
$u = v - at$
$u = 18 - (2.7)(5.4) = 18 - 14.58 = 3.42\ \text{m/s}$

Step2: Calculate distance traveled

Use $s = ut + \frac{1}{2}at^2$
$s = (3.42)(5.4) + \frac{1}{2}(2.7)(5.4)^2$
$s = 18.468 + \frac{1}{2}(2.7)(29.16)$
$s = 18.468 + 39.324 = 57.792\ \text{m}$

Step1: Define known values

Take upward as positive: initial velocity $u=25\ \text{m/s}$, final velocity $v=-15\ \text{m/s}$ (downward), acceleration $a=-9.8\ \text{m/s}^2$

Step2: Solve for time

Use $v = u + at$, rearrange for $t$:
$t = \frac{v - u}{a}$
$t = \frac{-15 - 25}{-9.8} = \frac{-40}{-9.8} \approx 4.08\ \text{s}$

Answer:

58 m

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Question 9