PHYSICS
Chp. 4 – Homework 3
Answer Key
Concept Review
5.
a.
v
t
b. The car’s acceleration is equal to the slope of the line graph.
Problems
2.
a.
b.
speed of car A = 50 m/s; speed of car B = 30 m/s
xa = v ∙ t = (30 m/s)(60 s) = 1800 m
xb = xo + vo(t) + (1/2)(a∙t2)
xb = 0 + 0 + 0.5 (1 m/s2)(60 s)2 = 1800 m
-or- you could find the area under the curve, instead of using equations
c.
d.
e.
car B is constant from 0-70 s; car A is constant from 60-70 s.
The slope of each line will give you the acceleration.
aa = 1 m/s2
ab = 0 m/s2
Use the equations or area under the curve.
Choice 2 gives the greatest distance traveled by the car.
3.
a = ∆v / ∆t
v = (0.40 m/s2)(12 s) = 4.8 m/s
4.
a.
b.
vavg = ∆x / ∆t = 50 m / 2 s = 25 m/s
1
𝑥 = 𝑣0 ∙ 𝑡 + 2 ∙ 𝑎 ∙ 𝑡 2
50 = 0.5(𝑎)(2 𝑠)2
a = 25 m/s2
𝑣 = 𝑣0 + 𝑎 ∙ 𝑡
v = (25 m/s2)(2 s) = 50 m/s
c.
a = 25 m/s2
5.
This problem is not logical, since the object is launched away from Earth and has a
positive velocity and acceleration.
But if you had to solve it, use:
t = v/a to get time, then use y = ½(a)(t2) to get the distance.
6.
a.
b.
c.
acceleration is the slope.
= (59 m/s) / (6 s) = 9.8 m/s2
1
𝑥 = 2 (𝑣 + 𝑣0 )𝑡
x = 0.5(39 m/s + 78 m/s)(4 s) = 234 m
fastest way to solve, use slope:
a = 0.33 m/s2
Could use the equation 𝑣 = 𝑣0 + 𝑎 ∙ 𝑡
The answers may vary, since the graph in the book is not good.
7.
Area under the first curve (area of the left rectangle).
Remember to convert minutes to hours.
distance = 5 miles
8.
A-B: cart accelerates away from the garage
B-C: cart travels with constant velocity away from the garage
C-D: cart travels at a constant velocity away from the garage that has a lower velocity
than from B-C.
D-E: cart is stopped
E-F: the cart accelerates towards the garage (hard to tell from the graph)
F-G: cart travels towards the garage at a constant velocity
G-H: the cart travels towards the garage with a negative acceleration
9.
graph A (remember it’s speed vs. time, not velocity vs. time, so direction does not matter.
All that matters is that the ball’s speed is increasing).
11.
𝑣 = 𝑣0 + 𝑔 ∙ 𝑡
v =114 m/s + (-9.8 m/s2)(8 s)
= 35.6 m/s in the up direction
12.
𝑦 = 𝑦0 + 𝑣0 ∙ 𝑡 + 2 ∙ 𝑔 ∙ 𝑡 2
y = 0 + (58 m/s)(7.2 s) + ½(-9.8 m/s2)(7.2 s)2
y = 170 m
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