All Solutions
Page 96: Lesson Check
For this part, we are given that the position-time graphs of two objects are a straight line and a parabola. The straight line position-time graph has a constant velocity, hence 0 acceleration. The parabolic position-time graph has a velocity that changes linearly, hence it has a constant acceleration. The graph corresponding to the constant acceleration is the $textbf{graph of a parabola}$.
In this problem, the acceleration of a car changes from positive to negative. We see how this affects the shape of the position-time graph. The sign of the acceleration corresponds to the curvature of the position time graph. If the curvature is upwards (like a cup), the acceleration is positive. If the curvature is downwards, like a cap, the acceleration is negative. Hence, in the given scenario, the position-time graph would $textbf{change curvature from upward to downward.}$
For this part, we are given four position-time graphs and we how the motion of the two objects differ.
For this graph, both objects $A$ and $B$ have constant slope – they both have constant velocity. However, the slope for $A$ is greater and more positive than the slope of $B$, so
$$
boxed{v_{A} > v_{B}}
$$
In this graph, $A$ has a parabolic upwards position-time graph, and $B$ has a linear. This means that $A$ must have a positive acceleration, while $B$ has zero acceleration,
$$
boxed{a_{A} > 0~mathrm{and }~a_{B} = 0}
$$
In this graph, both of the position-time graphs are parabolic upwards, so their accelerations are positive. Object $A$ has a steeper rise, so it must have a greater acceleration
$$
boxed{a_{A} > a_{B} > 0}
$$
In this graph, object $A$ has parabolic upwards, and object $B$ has parabolic downwards. This means that $A$ has a positive acceleration, and $B$ has a negative acceleration.
$$
boxed{a_{A} > 0 > a_{B}}
$$
textbf{Concept:}
$$
With the help of given data we will develop the position-time equation for the ball, assuming the ground to be $x=0$ and upward to be the positive direction
$$
textbf{Solution:}
$$
$$
x_{f}=x_{i}+v_{i}t+frac{1}{2}at^2=(1.5m)+(3.0m/s)t+frac{1}{2}(-9.81m/s^2)t^2
$$
$$
color{#4257b2} boxed{bf x_{f}=(1.5m)+(3.0m/s)t-(4.9m/s^2)t^2}
$$
x_{f}=(1.5m)+(3.0m/s)t-(4.9m/s^2)t^2
$$
The friend has a constant speed. The friend’s position-time equation must be
$$
begin{align*}
x_text{f, 2} &= v_{2}t \
x_text{f, 2} &= left( 3.5~mathrm{m/s} right)t
end{align*}
$$
$$
begin{align*}
x_text{f, 1} &= frac{1}{2}a_{1}left( t – t’ right)^{2} \
x_text{f, 1} &= frac{1}{2} left( 2.4~mathrm{m/s^{2}} right) left( t – 2.0~mathrm{s} right)^{2}
end{align*}
$$
$$
begin{align*}
x_text{f, 1} &= x_text{f, 2} \
frac{1}{2} left( 2.4~mathrm{m/s^{2}} right) left( t – 2.0~mathrm{s} right)^{2} &= left( 3.5~mathrm{m/s} right)t \
implies 0 = frac{1}{2} left( 2.4~mathrm{m/s^{2}} right) t^{2} + &left[ left( frac{1}{2} left( 2.4~mathrm{m/s^{2}} right) left( -2 right) left( 2.0~mathrm{s} right) right) – left( 3.5~mathrm{m/s^{2}} right) right]t + frac{1}{2} left( 2.4~mathrm{m/s^{2}} right) left( 2.0~mathrm{s} right)^{2} \
0 &= left( 1.2~mathrm{m/s^{2}} right)t^{2} + left( -8.3~mathrm{m/s} right)t + 4.8~mathrm{m} tag{1}
end{align*}
$$
$$
begin{align*}
t &= -frac{-8.3~mathrm{m/s}}{1.2~mathrm{m/s^{2}}} pm frac{sqrt{ left( -8.3~mathrm{m/s} right)^2 – 4left( 1.2~mathrm{m/s^{2}} right) left( 4.8~mathrm{m} right) }}{1.2~mathrm{m/s^{2}}} \
t &= 0.63697~mathrm{s}, 6.27967~mathrm{s}
end{align*}
$$
We take the time that is greater than 2, so they must have met at time
$$
boxed{ t = 6.3~mathrm{s} }
$$
We substitute this time to either of the position-time equation. On her friend, this must have been
$$
begin{align*}
x_text{f} &= v_{2}t \
&= left( 3.5~mathrm{m/s} right) left( 6.3~mathrm{s} right) \
&= 22.05~mathrm{m} \
x_text{f} &= boxed{ 22~mathrm{m} }
end{align*}
$$
Remember that there is a time delay for the cyclist. The velocity-time equation must be
$$
begin{align*}
v_text{f, 1} &= x_text{i, 1} + a_{1} left( t – t’ right) \
&= 0 + left( 2.4~mathrm{m/s^{2}} right) left( 6.3~mathrm{s} – 2.0~mathrm{s} right) \
&= 10.32~mathrm{m/s} \
v_text{f, 1} &= boxed{ 10.~mathrm{m/s} }
end{align*}
$$
item [a)] $t = 6.3~mathrm{s}$
item [b)] $x_text{f} = 22~mathrm{m}$
item [c)] $v_text{f, 1} = 10.~mathrm{m/s}$
end{enumerate}
The red graph is a graph of constant velocity. It was given that $textbf{motorcycle 1}$ has constant velocity.
We first use the position-time equation of motorcycle 1
$$
begin{align*}
x_text{f} &= x_text{i, 1} + v_{1}t \
&= 50.0~mathrm{m} + left( 6.50~mathrm{m/s} right) left( 10.0~mathrm{s} right) \
x_text{f} &= boxed{ 115~mathrm{m} }
end{align*}
$$
Now, we use the position-time equation of motorcycle 2
$$
begin{align*}
x_text{f} &= x_text{i, 1} + v_{2}t + frac{1}{2}a_{2}t^{2} \
x_text{f} &= 0 + 0 + frac{1}{2}a_{2}t^{2} \
implies a_{2} &= frac{2x_text{f}}{t^{2}} \
&= frac{2 left( 115~mathrm{m} right)}{left( 10.0~mathrm{s} right)^{2}} \
a_{2} &= boxed{ 2.30~mathrm{m/s^{2}} }
end{align*}
$$
Now, for the velocity-time equation of motorcycle 2
$$
begin{align*}
v_text{f, 2} &= v_text{i, 2} + a_{2}t \
&= 0 + left( 2.30~mathrm{m/s^{2}} right) left( 10.0~mathrm{s} right) \
v_text{f, 2} &= boxed{ 23.0~mathrm{m/s} }
end{align*}
$$
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