speed - distance catching up

• Feb 5th 2008, 02:13 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: $\displaystyle 35ms^{-1}$; police car acceleration: $\displaystyle 4ms^{-2}$

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

The distance travelled by the motorbike is given by

$\displaystyle d_m=35t$

The distance travelled by the police car can be expressed by

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

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

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

Thankyou!
• Feb 5th 2008, 07:50 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: .$\displaystyle v_m \:=\:35$
. . Integrating: .$\displaystyle d_m \:=\:35t + C_1$
At $\displaystyle t = 0,\:d = 0\quad\Rightarrow\quad C_1 = 0$
. . Hence: .$\displaystyle \boxed{d_m \:=\:35t}$

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

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

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

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

The police car overtakes the motorbike in 17½ seconds.

The police car travels the same distance as the motorbike.

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