Question

1. josselin drops a coin off a bridge and the coin falls at 9.8 m/s² for 2.6 s. what is the velocity of the coin the instant before it hits the water below?

a = 9.8 m/s²
δv =
t = 2.6 s

1. a car travels from rest to a speed of 39 m/s in 10 s. what is the cars acceleration in m/s²?

a =
δv =
t =

1. determine the total distance & total displacement in the following image.

total distance:
total displacement:
study guide - unit 1 motion

Explanation:

Step1: Identify the kinematic - equation for question 38

The kinematic equation $v = v_0+at$. Since the coin is dropped, $v_0 = 0$, $a = 9.8\ m/s^2$, and $t = 2.6\ s$.
$v=0 + 9.8\times2.6$

Step2: Calculate the velocity

$v=9.8\times2.6=25.48\ m/s$

Step3: Identify the kinematic - equation for question 39

The acceleration formula is $a=\frac{\Delta v}{t}$. The car starts from rest, so $v_0 = 0$, $v = 39\ m/s$, and $t = 10\ s$. Then $\Delta v=v - v_0=39 - 0 = 39\ m/s$.
$a=\frac{39}{10}$

Step4: Calculate the acceleration

$a = 3.9\ m/s^2$

Step5: Define distance and displacement for question 40

Distance is the total length of the path traveled, and displacement is the straight - line distance from the initial point A to the final point B. Without specific values for the path length and the straight - line distance in the figure, we assume the path length (distance) is $d$ and the straight - line distance (displacement) is $s$. If we measure the path length and the straight - line distance, the total distance is the sum of all the lengths of the segments of the curved path, and the displacement is the length of the straight line from A to B.

Answer:

1. $25.48\ m/s$
2. $3.9\ m/s^2$
3. Total Distance: Measure of the curved path length; Total Displacement: Measure of the straight - line distance from A to B