Question

A speeder passes a parked police car at a
constant speed of 28.4 m/s. At that instant,
the police car starts from rest with a uniform
acceleration of 2.93 m/s2.
How much time passes before the speeder
is overtaken by the police car?
Answer in units of s

Answer 1

Let's set x = 0 where the police car is. What's the equation of motion for the two cars?
\[ x_{speeder}(t) = ... what? \]
\[ x_{police}(t) = ... what? \]

Answer 2

the speeder's location xs(t) is
xs(t) = 28.4t

Now what's the police car's position, xp(t) ?

Answer 3

(1/2)(2.93)t^2

Answer 4

Right, exactly. Now when are the two equal?

Answer 5

set them = to each other?

Answer 6

when the police car catches the speeder

Answer 7

Set them equal and find the value of t for which that is the case.

Answer 8

48.46 and 0

Answer 9

obviously its not 0

Answer 10

Right. t = 0 is when the driver first goes past the police car. You want the other answer.

Answer 11

and it is not correct. I've already tried it previously.

Answer 12

t = 48 sec isn't right. Check your calculation

Answer 13

\[ 28.4 t = \frac{2.93}{2} t^2 \]

Answer 14

\[ t = 0 \]
or
\[ t = \frac{2 \times 28.4}{2.93} \]

Answer 15

I made an error somewhere, thank-you.

Answer 16

2.93? wait

Answer 17

2 * a
a=.586?

Answer 18

.586^2 ?

Answer 19

noooo

Answer 20

1/2 * 2.93?

Answer 21

The position of the speeder is\[ x_s(t) = 28.4 t \]The position of the police car is
\[ x_p(t) = \frac{1}{2} 2.93t^2 \]

Hence the positions are equal when
\[ 28.4 t = \frac{1}{2}2.93t^2 \]
Now solve for t as I've written above. You can factor out a t from both sides if t is not zero, which gives you
\[ \frac{1}{2}2.93t = 28.4 \]
hence
\[ t = \frac{2 \times 28.4}{2.93} \]
Now simplify that expression.

Answer 22

19.39

Answer 23

Yes, t = 19.39 seconds

Answer 24

I was making this more difficult by using quadratic equation. Thank you for the help and clarification.