How Far Ahead Of The Police Car Was The Speed Car When The Chase Began?
Problem set number 1- Palak Sharma
Distance a scalar quantity that refers to “how much ground an object has covered” during its motion.
Formulas used:
Newtons laws are: (1) Every object moves in a straight line unless acted upon by a force. (2) The acceleration of an object is directly proportional to the net force exerted and inversely proportional to the object's mass. (3) For every action, there is an equal and opposite reaction.
Formulas used:
F=ma
Displacement is a vector quantity that refers to it is the object’s overall change in position.
Formulas used:
Speed is a scalar quantity, is the rate at which an object covers distance
Formulas used:
S=d/t
Velocity is a vector quantity, and average velocity can …show more content…
Formulas used:
A=v2-v1/t
Friction word that refers to any force that resists relative opposite motion
Formulas used:
Ff=uFn
Forces something that causes a change in the motion of an object
Formulas used :
F=ma
3. Given: the graph
Required: a. How far ahead of the police car was the speeding car when the chase began?
b. What was the speeding car's velocity during the first 14 s of the chase?
c. How far did the speeding car travel before stopping?
d. What was the average velocity of the police car during the pursuit?
e. What was the police car's velocity at t = 10.0 s?
f. If the police car started from rest, what was its acceleration (assume it to be constant) during this chase?
Analysis: see solutions below
Solutions:
a. Reading from the graph:
Therefore the speeding car was 50 meter ahead of the police car when the race began.
b. vavg=d/t
= 415m-50m/14s
=26m/s
Therefore average velocity of the car is 26m/s during the first 14 seconds
c. 415-40
=70 m
Therefore the traveled 370m before stopping
d. v= d/t
=405m/ 18s
= 22.5 …show more content…
Given:
Weight: 3000 kg
Required:
a. Draw a free-body diagram of the elevator and calculate the values of all the forces that are acting on it when at rest.
b. If the elevator is ascending at a speed of 3.0 m/s, what are the values of the forces acting at this point?
c. If the elevator is descending at 3.0 m/s2, what are the values of all the forces acting at this point?
Analysis: see solutions below
Solutions:
a.
Fg=mg Fg = mg = (3000 kg)(9.8 N/kg)
= 30000 N up
Therefore it is 30000 N up
b. the elevator is moving with constant speed Fnet= 0 and that means acceleration is 0 therefore it is the same as part a (30000N downwards and upwards
c. let up be positive and down be negative
Downward acceleration= Net force/m= (mg-T)/m a = g-T/m
T= (g-a) m = (9.8-3)3000
= 20400N (Up)
Net Force acting upwards
= mg
= 3000x 9.8
= 29400N (Down)-9000N
=20400N
Therefore the net force acting upwards is 20400N and the net force acting downwards is 20400N.
9. Given: mass = 45 kg mK= Coefficient of Kinetic Friction = 0.18 d = 2.0m
Fnet = 50 N
Vf = 1.0 m/s
Required:
V1=?
Analysis: see solutions below
Solution:
F=ma
a=fa-umg/m
=50n- 0.18(45kg)(9.8m/s^2) /
Distance a scalar quantity that refers to “how much ground an object has covered” during its motion.
Formulas used:
Newtons laws are: (1) Every object moves in a straight line unless acted upon by a force. (2) The acceleration of an object is directly proportional to the net force exerted and inversely proportional to the object's mass. (3) For every action, there is an equal and opposite reaction.
Formulas used:
F=ma
Displacement is a vector quantity that refers to it is the object’s overall change in position.
Formulas used:
Speed is a scalar quantity, is the rate at which an object covers distance
Formulas used:
S=d/t
Velocity is a vector quantity, and average velocity can …show more content…
Formulas used:
A=v2-v1/t
Friction word that refers to any force that resists relative opposite motion
Formulas used:
Ff=uFn
Forces something that causes a change in the motion of an object
Formulas used :
F=ma
3. Given: the graph
Required: a. How far ahead of the police car was the speeding car when the chase began?
b. What was the speeding car's velocity during the first 14 s of the chase?
c. How far did the speeding car travel before stopping?
d. What was the average velocity of the police car during the pursuit?
e. What was the police car's velocity at t = 10.0 s?
f. If the police car started from rest, what was its acceleration (assume it to be constant) during this chase?
Analysis: see solutions below
Solutions:
a. Reading from the graph:
Therefore the speeding car was 50 meter ahead of the police car when the race began.
b. vavg=d/t
= 415m-50m/14s
=26m/s
Therefore average velocity of the car is 26m/s during the first 14 seconds
c. 415-40
=70 m
Therefore the traveled 370m before stopping
d. v= d/t
=405m/ 18s
= 22.5 …show more content…
Given:
Weight: 3000 kg
Required:
a. Draw a free-body diagram of the elevator and calculate the values of all the forces that are acting on it when at rest.
b. If the elevator is ascending at a speed of 3.0 m/s, what are the values of the forces acting at this point?
c. If the elevator is descending at 3.0 m/s2, what are the values of all the forces acting at this point?
Analysis: see solutions below
Solutions:
a.
Fg=mg Fg = mg = (3000 kg)(9.8 N/kg)
= 30000 N up
Therefore it is 30000 N up
b. the elevator is moving with constant speed Fnet= 0 and that means acceleration is 0 therefore it is the same as part a (30000N downwards and upwards
c. let up be positive and down be negative
Downward acceleration= Net force/m= (mg-T)/m a = g-T/m
T= (g-a) m = (9.8-3)3000
= 20400N (Up)
Net Force acting upwards
= mg
= 3000x 9.8
= 29400N (Down)-9000N
=20400N
Therefore the net force acting upwards is 20400N and the net force acting downwards is 20400N.
9. Given: mass = 45 kg mK= Coefficient of Kinetic Friction = 0.18 d = 2.0m
Fnet = 50 N
Vf = 1.0 m/s
Required:
V1=?
Analysis: see solutions below
Solution:
F=ma
a=fa-umg/m
=50n- 0.18(45kg)(9.8m/s^2) /