Question

Amir starts riding his bike up a 200-m-long slope at a speed of 18 km/h, decelerating at 0.20 m/s² as he goes up. At the same instant, Becky starts down from the top at a speed of 6.0 km/h, accelerating at 0.40 m/s² as she goes down. How far has Amir ridden when they pass?

Short answer

The distance traveled by Amir when Becky and Amir passes is 62 m

Step-by-step solution

Step 1. Write the given information

The velocity of Amir is $v_A=18km/hr=5m/sec$

the acceleration of Amir is$a_A=-0.2m/s^2$

The distance of the slope is$S=200m$

The velocity of Becky is $v_B=6km/h=1.6m/sec$

The acceleration is$a_B=0.4m/s^2$

Step 2. To determine the time at which they will cross each other

Let the distance traveled by Amir at the point of intersection is $s_1$
The time at this instant is $t$

Write the equation of motion for Amir

$$\begin{gathered} s_1=v_{A}t+\frac{1}{2}{a}_A{t}^2 \\ {s}_1=5t+\frac{1}{2}(-0.2)t^2 \\ {s}_1=5t-0.1t^2.......\left(1\right) \end{gathered}$$

Similarly, let the distance traveled by Becky at the point of intersection is $s_2$

Write the equation of motion for Becky

$$\begin{gathered} s_2=v_{B}t+\frac{1}{2}{a}_B{t}^2 \\ {s}_2=1.6t+\frac{1}{2}(0.4)t^2 \\ {s}_2=1.6t+0.2t^2.....\left(2\right) \end{gathered}$$

Since the total distance is given by $S=200m$
therefore, $s_1+s_2=S$
Now add equations (1) and (2)
$$\begin{gathered} s_1+s_2=5t-0.1t^2+1.6t+0.2t^2 \\ 200=6.6t+0.1t^2 \\ 2000=66t+t^2 \\ {t}^2+66t-2000=0 \end{gathered}$$

Using the quadratic equation formula,

$t=22.5sec$

Therefore, Amir and Becky cross each other after time 22.5 sec

Step 3. To determine the distance traveled by Amir when they pass each other

Substitute the value of time t in the equation (1)

$$\begin{gathered} s_1=5(22.5)-0.1{(22.5)}^2 \\ {s}_1=112.5-50.6 \\ {s}_1=62m \end{gathered}$$

Therefore, the distance traveled by Amir when Becky and Amir passes is 62 m