Assume the motion when you are in the car or in the school bus to go to the school.
To describe the motion the first thing you need is a point of reference. Assume this is your house.
This should be a description:
- When you are sitting and the car has not started to move you are at rest.
- The car starts moving from rest, gaining speed, accelerating. You start to move away from your house, with a positive velocity (from you house to your school) and positive acceleration (velocity increases).
- The car reaches a limit speed of 40mph, and then moves at constant speed. The motion is uniform, the velocity is constant, positive, since you move in the same direction), and the acceleration is zero.
- When the car approaches the school, the driver starts to slow down. Then, you speed is lower but yet the velocity is positive, as you are going in the same direction. The acceleration is negative because it is in the opposite direction of the motion.
- When the car stops, you are again at rest: zero velocity and zero acceleration.
- In all the path your velocity was positive, constant at times (zero acceleration) and variable at others (accelerating or decelerating).
- When you comeback home, then you can start to compute negative velocities, as you will be decreasing the distance from your point of reference (your house).
If Juan used a Celsius thermometer, it would tell him the Celsius temperature.
If he added 273 to that number, he'd have the "absolute" or Kelvin temperature.
Answer:
a) x(t) = 10t + (2/3)*t^3
b) x*(0.1875) = 10.18 m
Explanation:
Note: The position of the horse is x = 2m. There is a typing error in the question. Otherwise, The solution to cubic equation holds a negative value of time t.
Given:
- v(t) = 10 + 2*t^2 (radar gun)
- x*(t) = 10 + 5t^2 + 3t^3 (our coordinate)
Find:
-The position x of horse as a function of time t in radar system.
-The position of the horse at x = 2m in our coordinate system
Solution:
- The position of horse according to radar gun:
v(t) = dx / dt = 10 + 2*t^2
- Separate variables:
dx = (10 + 2*t^2).dt
- Integrate over interval x = 0 @ t= 0
x(t) = 10t + (2/3)*t^3
- time @ x = 2 :
2 = 10t + (2/3)*t^3
0 = 10t + (2/3)*t^3 + 2
- solve for t:
t = 0.1875 s
- Evaluate x* at t = 0.1875 s
x*(0.1875) = 10 + 5(0.1875)^2 + 3(0.1875)^3
x*(0.1875) = 10.18 m
