# How does the object at constant speed pass the accelerating object?

A car starts from rest and move with constant acceleration of $4\frac{m}{s^2}$ and at the same time a motorcyclist moving with a constant speed of $36\frac{km}{hr}$ overtakes and passes the car. Find

a) How far beyond the starting point will the car overtakes the motorcyclist?

b) What will be the speed of the car at the time when it overtakes the motorcycle?

I am confused by part a of the question. Can anyone tell me, if the motorcyclist had a constant speed of $36\frac{\mathrm{km}}{\mathrm{hr}}$ and the car had a constant acceleration or $4\frac{\mathrm{m}}{\mathrm{s}^2}$, how the bike overtook and passed by the car so quickly? Also it is not clear here from where the bike started?

• What do you think A car starts...at the same time a motorcyclist...passes the car indicates for the starting positions? – Kyle Kanos Feb 26 '15 at 16:34
• Well if I agree with you then my problem is still active that is saying that how can motorcyclist overtake car and pass it so fast when it's speed is lesser then that of the car? – user73555 Feb 26 '15 at 16:37
• The car starts from rest and accelerates. The motorcycle moves at constant velocity. Draw a picture, it should help immensely. – Kyle Kanos Feb 26 '15 at 16:38
• I just don't understand the issue you're having. Clearly the car & bike have different velocities initially and they start at the same point. I fail to see how you think the bike wouldn't pass the car immediately. – Kyle Kanos Feb 26 '15 at 16:49
• I am not sure I follow or how you got that from the problem statement. A car starts from rest What does that mean for its velocity? And how does that compare to $v_{bike}=36\,\rm m/s$? – Kyle Kanos Feb 26 '15 at 16:53

Let's say car and bike be at rest at $1pm$ so, $v_c=0$ and $v_b=0$.

Calculations for motion of car:

Since car is moving with constant acceleration,

At 1:00:00pm, $v_c=0m/s$, $S_c=0m$

At 1:00:01pm, $v_c=4m/s$, $S_c=4m$

At 1:00:02pm, $v_c=8m/s$, $S_c=12m$

At 1:00:03pm, $v_c=12m/s$, $S_c=24m$

At 1:00:04pm, $v_c=16m/s$, $S_c=40m$

Calculations for motion of bike:

Since bike is moving with constant speed,

At 1:00:00pm, $v_b=0$, $S_b=0m$

At 1:00:01pm, $v_b=10m/s$, $S_b=10m$

At 1:00:02pm, $v_b=10m/s$, $S_b=20m$

At 1:00:03pm, $v_b=10m/s$, $S_b=30m$

At 1:00:04pm, $v_b=10m/s$, $S_b=40m$

So from above calculations that in $4$ $seconds$ your car will overtake the bike after covering the distance of $40$ $meters$.

The motorcycle is moving at $10\frac{m}{s}$, so his distance from a starting point of 0 each second can be modeled as:

$$y = 10x$$

The car is moving at a constant acceleration, so that distance from an origin of 0 can be modeled as:

$$y = 2(x^2 + x)$$

Therefore, when they are equal to each other, they are at equal distances from the starting point.

$$10x = 2(x^2 + x) = 4$$

At 4 seconds from the starting point, they are at the same distance from the starting point, so when time is greater than 4 seconds, the car has overtaken the motorcycle.

Speed is the absolute value of the derivative of the distance over time equation.

$$|f'(x) = 4x + 2|$$

Plug in 4, and you see that the speed of the car at the 4 second mark is $18\frac{m}{s}$.

Basically, we give the motorcycle and the car an arbitrary start point. 0 is usually easiest. The motorcycle was already traveling his constant speed before we starting measuring the acceleration of the car, but for the sake of the calculation, we start our measurements at the point where the car begins its initial acceleration.

The car doesn't reach the motorcycle until 4 seconds after it starts from a dead stop. Keep in mind, the equation tries to throw you off by showing the acceleration of the car and speed of the motorcycle in different factors of the same units.

If you translate the units, you'll see that the car is accelerating at a rate of $14.4\frac{km}{hr}$ every second. Looking at it this way, it is easy to see why the car catches the motorcycle so quickly.