ACCELERATION
Acceleration:
Q 47, Pg 60) After landing, a jetliner on a straight runway taxis to a stop at an average velocity of -35 km/h. If the plane taken 7s to come to rest, what are the plane's initial velocity and acceleration?
A:
Initial Velocity
Vavg= (Vi+Vf)/2
35 km/h=(X+0)/2
Using algebra to solve for X,
Vi= -70km/h which equals to -19.44m/s
Acceleration
Change in velocity of plane (Vf-Vi) / time (7s)
(0-19.44) / 7
=2.78m/s^2
Q 47, Pg 60) After landing, a jetliner on a straight runway taxis to a stop at an average velocity of -35 km/h. If the plane taken 7s to come to rest, what are the plane's initial velocity and acceleration?
A:
Initial Velocity
Vavg= (Vi+Vf)/2
35 km/h=(X+0)/2
Using algebra to solve for X,
Vi= -70km/h which equals to -19.44m/s
Acceleration
Change in velocity of plane (Vf-Vi) / time (7s)
(0-19.44) / 7
=2.78m/s^2
V-T Graph
![Picture](/uploads/1/7/0/4/17041910/315529.jpg)
Explanation: The problem states that the plane taxis to a stop at an average velocity of -35 km/h. This graph shows that the plane starts out at -70km/h on the runway and slowly decreases (has a negative slope) and reaches 0 m/s in 7 seconds--at a constant rate. The average velocity can be calculate--> (Vi+Vf)/2-->(70km/h+0km/h) /2 = -35 km/h.
A-T Graph
![Picture](/uploads/1/7/0/4/17041910/639688.jpg)
Explanation: The acceleration is negative because the slope of the v-t graph is negative and because the plane has a constant negative acceleration in order to come to a stop. The acceleration is constant because the plane constantly travels at -35km/h (velocity) and acceleration is the change in velocity per unit of time. (Acc. is negative because the plane is slowing down).
P-T Graph
![Picture](/uploads/1/7/0/4/17041910/4584143.jpg)
Explanation: The p-t graph is curved because the velocity (shown by slope) decreases with every second that passes. The p-t graph ends flat at 7s because thats when the plane stops moving-no velocity.