Consider a toy rocket that undergoes one dimensional motion. It is launched from
ID: 1572687 • Letter: C
Question
Consider a toy rocket that undergoes one dimensional motion. It is launched from rest straight up from the ground and rises under constant acceleration for t1 seconds, where it loses power. It hits the earth t2 seconds later. You can consider t1 = t1 – t0 and t2 = t2 – t1. What was the toy rocket’s acceleration before t1? Hw 2 written problems 1. Consider a toy rocket that undergoes one dimensional motion. It is launched from rest straight up from the ground and rises under constant acceleration for t seconds, where it loses power. It hits the earth seconds later. You can consider Ati-t1-to and At-t-t. what was the toy rocket's acceleration before t,? Ch2 p69. A cat is sleeping on the floor in the middle of a 3.0 m wide room when a barking dog enters with a speed of 1.50 m/s. As the dog enters, the cat immediately accelerates at 0.85 m/s toward an open window on the opposite side of the room. The dog is a bit startled by the cat and begins to slow down at 0.10 m/s as soon as it enters the room. How far is the cat in front of the dog as it leaps through the window?Explanation / Answer
for the cat :
vo = initial velocity = 0 m/s
a = acceleration = 0.85 m/s2
x = distance travelled to reach the window = 3/2 = 1.5 m
t = time of travel
using the equation
x = vo t + (0.5) a t2
1.5 = 0 t + (0.5) (0.85) t2
t = 1.88 sec
consider the motion of dog :
vo = initial velocity = 1.5 m/s
a = acceleration = - 0.1 m/s2
x = distance travelled by the dog = ?
t = time of travel = 1.88 sec
using the equation
x = vo t + (0.5) a t2
x = (1.5 x 1.88) + (0.5) (- 0.1) (1.88)2
x = 2.64 m
distance between cat and dog is given as
distance = width of room - distance travelled by dog = 3 - 2.64 = 0.36 m