Racing Toy Cars

Classical Mechanics Level 1

A small toy car (car A) is kicked so that it travels at $2 \text{ m}/\text{s}$ to the right without friction. Simultaneously, another toy car (car B) accelerates from rest via an electric motor at $0.9 \text{ m}/\text{s}^2$ to the right without friction. Which reaches the finish line $10 \text{ m}$ away first?

Car A Car B The two cars tie. Not enough information

1 solution

Matt DeCross
Apr 30, 2016

Car A takes:

$t = \frac{d}{v} = \frac{10 \text{ m}}{2 \text{ m}/\text{s}} = 5 \text{ s}$

to reach the finish line.

Since car B is constantly acclerating, its velocity over time is given by $v= v_0 + at$ by integrating. Integrating again, one obtains the position as a function of time:

$x= x_0 + v_0 t + \frac12 at^2.$

Since car B accelerates at rest and we only care about $x - x_0$ , the distance traveled, this can be rewritten as:

$d = \frac12 at^2.$

Solving for $t$ given $a = 0.9 \text{ m}/\text{s}^2$ and $d = 10 \text{ m}$ yields:

$t = \sqrt{20 / .9} \text{ s} \approx 4.7 \text{ s}.$

As $4.7 < 5$ , car B reaches the finish first.

Here another solution by its velocity we can assume it as ...

Car A traveled 2 m/s Car B traveled 0.9 m/s² so individul velocity we have Car A = Car B

2/60 m/min = 0.9/60×60 m/min 0.00025 m/min = 0.3333 m/min i.e. car A has traveled speed of 0.00025 m/min which is lower than Car B traveled speed of 0.3333 m/min. so we can assume which will reach first by its velocity ... 0.00025<0.3333 so car A takes much less traveled per minute compare to car B . so Car B will be reach first...

A Former Brilliant Member - 4 years, 9 months ago

Racing Toy Cars

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