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Question
acceleration practice worksheet (including negative acceleration)
use the equation:
a = (v_f - v_i)/t
or equivalently:
δv = a × t
show all work, include units, and circle your final answer. remember that acceleration can be negative when an object is slowing down.
solve for final velocity (v_f)
- a car starts at 28 m/s and accelerates at -3.0 m/s² for 4.0 seconds. what is its final velocity?
- a runner with an initial speed of 12 m/s accelerates at -1.5 m/s² for 6.0 seconds. what is the runner’s final speed?
solve for initial velocity (v_i)
- a train slows down at -2.5 m/s² for 8.0 seconds and ends at a final velocity of 10 m/s. what was its initial velocity?
- a motorcycle decelerates at -4.0 m/s² for 5.0 seconds and ends at 15 m/s. what was its starting velocity?
solve for acceleration (a)
- a bus goes from 30 m/s to 18 m/s in 4.0 seconds. what is the buss acceleration?
Step1: Recall the acceleration - velocity formula
The formula $a=\frac{v_f - v_i}{t}$ can be rewritten as $v_f=v_i + a\times t$ and $v_i=v_f - a\times t$ and $a=\frac{v_f - v_i}{t}$.
Step2: Solve problem 1
Given $v_i = 28\ m/s$, $a=- 3.0\ m/s^2$, $t = 4.0\ s$. Using $v_f=v_i + a\times t$, we substitute the values: $v_f=28+( - 3.0)\times4.0=28 - 12=16\ m/s$.
Step3: Solve problem 2
Given $v_i = 12\ m/s$, $a=-1.5\ m/s^2$, $t = 6.0\ s$. Using $v_f=v_i + a\times t$, we substitute the values: $v_f=12+( - 1.5)\times6.0=12 - 9 = 3\ m/s$.
Step4: Solve problem 3
Given $v_f = 10\ m/s$, $a=-2.5\ m/s^2$, $t = 8.0\ s$. Using $v_i=v_f - a\times t$, we substitute the values: $v_i=10-( - 2.5)\times8.0=10 + 20=30\ m/s$.
Step5: Solve problem 4
Given $v_f = 15\ m/s$, $a=-4.0\ m/s^2$, $t = 5.0\ s$. Using $v_i=v_f - a\times t$, we substitute the values: $v_i=15-( - 4.0)\times5.0=15 + 20=35\ m/s$.
Step6: Solve problem 5
Given $v_i = 30\ m/s$, $v_f = 18\ m/s$, $t = 4.0\ s$. Using $a=\frac{v_f - v_i}{t}$, we substitute the values: $a=\frac{18 - 30}{4.0}=\frac{-12}{4.0}=-3.0\ m/s^2$.
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