Question

chapter 3 linear motion
acceleration of free fall
a rock thrown from the top of a cliff picks up speed as it falls. pretend that a speedometer and odometer are attached to the rock to show readings of speed and distance at 1 - second intervals. both speed and distance are zero at time = zero (see sketch). note that after falling 1 second, the speed reading is 10 m/s and the distance fallen is 5 m. the readings for succeeding seconds of fall are not shown and are left for you to complete.
draw the position of the speedometer pointer and write in the correct odometer reading for each line. use ( g = 10 m/s^2 ) and neglect air resistance.
you need to know:
instantaneous speed of fall from rest: ( v = gt )
distance fallen from rest: ( d = v_{average}t ) or ( d=\frac{1}{2}gt^2 )

1. the speedometer reading increased by the same amount, ____ m/s, each second. this increase in speed per second is called ____
2. the distance fallen increases as the square of the ______
3. if it takes 7 seconds to reach the ground, then its speed at impact is ____ m/s, the total distance fallen is __ m, and its acceleration of fall just before impact is ____ ( m/s^2 ).

Explanation:

Step1: Calcular la aceleración

Dado que \( g = 10\ m/s^{2}\), la velocidad aumenta \( 10\ m/s\) cada segundo. Esto se debe a que la aceleración \( a = g\) y \( v=at\), donde \( a = 10\ m/s^{2}\) (aceleración).

Step2: Relación distancia - tiempo

De la fórmula \( d=\frac{1}{2}gt^{2}\), vemos que la distancia \( d\) es proporcional al cuadrado del tiempo \( t\).

Step3: Velocidad, distancia y aceleración para \( t = 7\ s\)

• Velocidad: \( v=gt\), con \( g = 10\ m/s^{2}\) y \( t = 7\ s\), entonces \( v=10\times7 = 70\ m/s\).
• Distancia: \( d=\frac{1}{2}gt^{2}=\frac{1}{2}\times10\times7^{2}=5\times49 = 245\ m\).
• Aceleración: La aceleración de la gravedad \( g\) es constante (se neglecta el aire), así que \( a = 10\ m/s^{2}\).

Answer:

1. \( 10\), aceleración.
2. tiempo.
3. \( 70\), \( 245\), \( 10\).