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Question
acceleration problems
$v_f = v_0+at$ $delta x = v_0t+\frac{1}{2}at^{2}$ $v_{f}^{2}=v_{0}^{2}+2adelta x$
- a car’s velocity increases from rest to 14 m/s, east, in 3.5 sec.
a. what is its acceleration?
b. what is its acceleration if it next slows down from 14 m/s to 7 m/s, east, in 2 sec?
- a mass is dropped from rest. what is its velocity after 5 sec if the acceleration due to gravity is -9.8m/s² straight down?
- a car accelerates from 6 m/s, west, to 20 m/s, west, over a distance of 50 m. what is its acceleration?
- a ball is dropped from rest from a roof 5 m above the ground. how long does it take to hit the ground if the acceleration is due to gravity and is -9.8 m/s² straight down?
Step1: Identify the formula for acceleration
The formula for acceleration is $a=\frac{v_f - v_0}{t}$, where $v_f$ is final - velocity, $v_0$ is initial - velocity, and $t$ is time.
Step2: Solve part a of question 1
Given $v_0 = 0$ m/s (starts from rest), $v_f=14$ m/s, and $t = 3.5$ s.
$a=\frac{v_f - v_0}{t}=\frac{14 - 0}{3.5}=4$ m/s².
Step3: Solve part b of question 1
Given $v_0 = 14$ m/s, $v_f = 7$ m/s, and $t = 2$ s.
$a=\frac{v_f - v_0}{t}=\frac{7 - 14}{2}=-3.5$ m/s².
Step4: Solve question 2
Use the formula $v_f=v_0+at$. Given $v_0 = 0$ m/s (starts from rest), $a=-9.8$ m/s², and $t = 5$ s.
$v_f=0+( - 9.8)\times5=-49$ m/s.
Step5: Solve question 3
Use the formula $v_f^{2}=v_0^{2}+2a\Delta x$. Given $v_0 = 6$ m/s, $v_f = 20$ m/s, and $\Delta x = 50$ m.
First, re - arrange the formula for $a$: $a=\frac{v_f^{2}-v_0^{2}}{2\Delta x}$.
$a=\frac{20^{2}-6^{2}}{2\times50}=\frac{400 - 36}{100}=\frac{364}{100}=3.64$ m/s².
Step6: Solve question 4
Use the formula $\Delta x=v_0t+\frac{1}{2}at^{2}$. Since $v_0 = 0$ m/s (starts from rest), the formula simplifies to $\Delta x=\frac{1}{2}at^{2}$.
Re - arrange for $t$: $t=\sqrt{\frac{2\Delta x}{a}}$. Given $\Delta x=-5$ m (downward) and $a=-9.8$ m/s².
$t=\sqrt{\frac{2\times(- 5)}{-9.8}}=\sqrt{\frac{10}{9.8}}\approx1.01$ s.
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- a. $4$ m/s²
b. $-3.5$ m/s²
- $-49$ m/s
- $3.64$ m/s²
- $1.01$ s