A Driver in a Blue Car Is Traveling at 50 Kmh Sees a Red Car

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Problems

$% latex2html id marker 848 $\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$$

A car travelling at a constant speed of 30 m/s passes a police car at rest. The policeman starts to move at the moment the speeder passes his car and accelerates at a constant rate of 3.0 m/ s ² until he pulls even with the speeding car. Find a) the time required for the policeman to catch the speeder and b) the distance travelled during the chase.

Solution:
We are given, for the speeder:

v ₀ ^s = 30 m/s	=	v ^s
a ^s = 0

and for the policeman:

v ₀ ^p	=
a ^p	=	3.0 m/s ².

a)

Distance travelled by the speeder x ^s = v ^s t = (30)t . Distance travelled by policeman x ^p = v ₀ ^p + ${\frac{1}{2}}$ a ^p t ² = ${\frac{1}{2}}$ (3.0)t ². When the policeman catches the speeder x ^s = x ^p or,

30t = $\displaystyle{\textstyle\frac{1}{2}}$ (3.0)t ².

Solving for t we have t = 0 or t = ${\frac{2}{3}}$ (30) = 20 s . The first solution tells us that the speeder and the policeman started at the same point at t = 0, and the second one tells us that it takes 20 s for the policeman to catch up to the speeder.

b)

Substituting back in above we find,
and,

x ^p = $\displaystyle{\textstyle\frac{1}{2}}$ (3.0)(20)² = 600 m = x ^s.

$% latex2html id marker 867 $\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$$

A car decelerates at

2.0 m/s ² and comes to a stop after travelling 25 m. Find a) the speed of the car at the start of the deceleration and b) the time required to come to a stop.

Solution:
We are given:

a	=	- 2.0 m/s ²
v	=
x	=	25 m

a): From v ² = v ₀ ² + 2ax we have v ₀ ² = v ² - 2ax = - 2(- 2.0)(25) = 100 m ²/s ² or v ₀ = 10 m/s.
b): From v = v ₀ + at we have t = ${\frac{1}{a}}$ (v - v ₀) = ${\frac{1}{-2.0}}$ (- 10) = 5 s.

$% latex2html id marker 878 $\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$$

A stone is thrown vertically upward from the edge of a building 19.6 m high with initial velocity 14.7 m/s. The stone just misses the building on the way down. Find a) the time of flight and b) the velocity of the stone just before it hits the ground.

Solution:
We are given,

v ₀	=	14.7 m/s
a	=	- 9.8 m/s ²

At the time the stone hits the ground,

a): From x = v ₀ t + ${\frac{1}{2}}$ at ² we have,
The two solutions are t = 4 s and t = - 1 s. The second (negative) solution gives the time the stone would have left the ground, and is unphysical in this case. The solution we want is the first one.
b): We substitute to find v = v ₀ + at = 14.7 - 9.8(4) = - 24.5 m/s. Note that the negative velocity correctly shows that the stone is moving down.

$% latex2html id marker 894 $\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$$

A rocket moves upward, starting from rest with an acceleration of

29.4 m/s ² for 4 s. At this time, it runs out of fuel and continues to move upward. How high does it go?

Solution:
For the first stage of the flight we are given:

v ₀	=
a	=	29.4 m/s ²
t	=	4 s

This gives, for the velocity and position at the end of the first stage of the flight: v ₁ = v ₀ + at = 29.4 m/s ²(4 s ) = 117.6 m/s and x ₁ = v ₀ t + ${\frac{1}{2}}$ at ² = ${\frac{1}{2}}$ (29.4)(4)² = 235.2 m.

For the second stage of the flight we start with,

v ₁	=	117.6 m/s
a	=	- 9.8 m/s ²

and end up with v ₂ = 0. We want to find the distance travelled in the second stage (x ₂ - x ₁). We have,
Therefore x ₂ = x ₁ + 705.6 = 235.2 + 705.6 = 940.8 m.

Next: About this document ... Up: Motion in One Dimension Previous: Freely Falling Bodies

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10/9/1997

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Source: https://theory.uwinnipeg.ca/physics/onedim/node9.html

A Driver in a Blue Car Is Traveling at 50 Kmh Sees a Red Car

Problems

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