# speed - distance catching up

• Feb 5th 2008, 03:13 AM
DivideBy0
speed - distance catching up
A speeding motorbike travels past a stationary police car. The police car starts accelerating immediately, and keeps accelerating until it has passed the bike.

DATA: motorbike speed: $35ms^{-1}$; police car acceleration: $4ms^{-2}$

How far does the police car travel before it overtakes the motorbike?

I tried working this out using the following method, but it is wrong. How come?

My Solution

Let the distance travelled by the motorbike be $d_m$ and the distance travelled by the police car be $d_p$. Let $t$ be time in seconds.

The distance travelled by the motorbike is given by

$d_m=35t$

The distance travelled by the police car can be expressed by

$d_p=4+8+...+t=\sum_{i=1}^t 4i = \frac{4t(t+1)}{2}=2t(t+1)$

When $d_p=d_m$, $35t=2t(t+1)$

Solving gives $t=16.5 \Rightarrow d = 577.5 \mbox{m}$

Thankyou!
• Feb 5th 2008, 08:50 AM
Soroban
Hello, DivideBy0!

Are we allowed to use Calculus?

Quote:

A speeding motorbike travels past a stationary police car. The police car
starts accelerating immediately, and keeps accelerating until it has passed the bike.

Data: motorbike speed: 35 m/s; police car acceleration: 4 m/s²

How far does the police car travel before it overtakes the motorbike?

We have: . $v_m \:=\:35$
. . Integrating: . $d_m \:=\:35t + C_1$
At $t = 0,\:d = 0\quad\Rightarrow\quad C_1 = 0$
. . Hence: . $\boxed{d_m \:=\:35t}$

We have: . $a_p \:=\:4$
. . Integrate: . $v_p \:=\:4t + C_2$
At $t=0,\:v = 0\quad\Rightarrow\quad C_2 = 0$
. . Hence: . $v_p \:=\:4t$

Integrate: . $d_p \:=\:2t^2 + C_3$
At $t = 0,\:d = 0\quad\Rightarrow\quad C_3 = 0$
.Hence: . $\boxed{d_p \:=\:2t^2}$

When are these two distances equal?
. . $2t^2 \:=\:35t\quad\Rightarrow\quad 2t^2-35t \:=\:0\quad\Rightarrow\quad t(2t-35) \:=\:0$

. . Hence: . $t \;=\;0,\:17.5$

The police car overtakes the motorbike in 17½ seconds.

The police car travels the same distance as the motorbike.

. . $d_p \;=\;d_m \;=\;35(17.5) \;=\;\boxed{612.5 \text{ m}}$