1. Explain the difference between displacement and distance 2.if it takes Matthe
ID: 2114026 • Letter: 1
Question
1. Explain the difference between displacement and distance
2.if it takes Matthew 10 hours and 13 minutes to go 925km from B. Hills sacramento, what is his average velocity?
a. using this example, explain what is meant by instantaneous speed and what i meant by average speed
3. list the motion equations. for each, list each variable, write what the variable means and the typical units for that variable in the MKS system
4. what is the reason for identifying all varibles before beginning a problem?
5.Ashley drives her car at an average velocity of 48.0km/hr,east; how long for her to drive 144 km?
6. suppose a treadmill has an average acceleration of 4.7 x10^-3m/s^2.
a. how much does its speed change after 5.0 minutes?
b. if the treadmill's initial speed is 1.7 m/s, what will its final speed be?
7.with an average acceleration of -0.50m/s2, how long will it take for a cyclist to stop a bicycle with an initial speed of 13.5 m/s?
8.A car accelerates uniformly from rest to a speed of 23.7 km/hr in 6.5 seconds. find the distance the car travels during that time.
9.michel puts on the breakes of his car which slows uniformly from 15.0 m/s to 0 m/s in 2.50 seconds. how many meters prior to the stop sign must he apply his brakes to stop in time?
10. A car enters the freeway at 6.4 m/s and accelerates uniformly for 3.2 km in 3.5 minutes. how fast is the car moving after this time.
11. tailor hits a volley ball with an initial velocity of 6.0 m/s straight upward. if the volleyball starts 2.0 m above the floor, how long will it be in the air before striking the floor?
12.A robot probe drops an object off a 239m cliff on Mars. gMars=3.7m/s2
a. find the time required for he object to reach the ground
b. what is the velocity when it hits the ground
12. what is the weight of a 58kg sign?
13. A 125 kg sign is suspended over the sidewalk with a boom and a cable. if the boom is stiking out from the building and th cable makes an angle of 57 degree with the brick wall of the building what is the tension in the cable?
b. with what force is the boom pushng the sign away from the building?
14. An airplane is flying at 250m/sec du east when it encounters a crosswind of 50m/sec due north.
a. calculate the resultant velocity of the plane
b. what correction should the pilot make in the cockpit in order to stay on the desired velocity?
Explanation / Answer
distance is the the magnitude of length covered irrespective of the direction but displacement is the shortest distance between two points and it also takes into condsideration the direction in which the diatance is travered
eq if u move in a circle of radius r
then distance = 2pir
displacement = 0 as u return to the starting point at last.
2) average velocity = total dsplacement /total time = 925 km/ 10.2167 = 90.53 km/hr
instanteneous speed is the speed at ant instant and the average speed is the average of all these instanteneous speed over the entire period.eg while travelling from point a to b we may not have a constant speed throughout so the soeed at anty particular instant will be the instanteneous speed and the average of all speed will give the average speed
3)
motion equations
v = final velocity ,m/s
u = initial velocity ,m/s
a = accelrration ,m/s^2
t = time ,s
S = dispalcement ,m
v = u + at
v^2 = u^2 + 2aS
S = ut + .5 at^2
4. to get an idea about what data we have and what we need to calculate and based on this information choose the equation that is most suitable
5.
time = displacement /velovity
= 144/48 = 3 hrs
6. suppose a treadmill has an average acceleration of 4.7 x10^-3m/s^2.
a)change in speed = v-u = at = 4.7 x 10^-3 x 5 x 60
= 1.41 m/s
b.
final speed = u + at = 1.7 + 1.41 = 3.11 m/s
8.
a = 6.58/6.5 = 1.01 m/s
S = .5 x a t^2 = .5 x 1 x 6.5^2 = 21.336 m
9.
a = 15/2.5 = 6 m/s^2
v = 0
u^2 - 2aS = 0
S = 18.75 m
10.
S = ut + .5 at^2
3200 = 6.4 x 3.5 x 60 + .5 x a x (3.5 x 60)^2
a = 0.084 m/s^2
11.
v = 0 at max height
6 - 9.8t = 0
t = 0.6122 s
this much time is taken to rise to the max height,same time is required to return to the starting position
now to fall a height of 2m with speed of 6 m/s
final velocity = v^2 = u^2 + 2aS
v^2 = 6^2 + 2 x 9.8 x 2
v = 8.67 m/s
v = u + at
t = 8.67 - 6 /9.8
t = 0.273s
total time 0.6122 + 0.6122 + .273 = 1.4974 s
12.
as u = 0
S = .5 x g t^2
t^2 = 239 x 2 / 3.7
t = 11.367 s
b.
v = at = 3.7 x 11.367 = 42.055 m/s
12. what is the weight of a 58kg sign?
weight = mg = 58 x 9.8 = 568.4 N
14. An airplane is flying at 250m/sec du east when it encounters a crosswind of 50m/sec due north.
a. calculate the resultant velocity of the plane
b. what correction should the pilot make in the cockpit in order to stay on the desired velocity?
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.