Sample Problems and Solutions
Early in Unit 6, quad kinematic equations were introduced and discussed. ONE useful problem-solving strategy was presented for use with save equations and two examples were given that illustrated the benefit away the our. Then, the application in the movement equations and the problem-solving strategy in free-fall motion was discussed plus illustrated. Inside diese part on Lesson 6, several sample problems will be presented. These problems allow any student of physics to test their understanding of the getting of the four kinematic equations to solve problems involving one one-dimensional motion of objects. They have encouraged to read each problem and practice the use of the strategy in to solution of an problem. Then click the button to check the ask or use the link at view the solution.
Check Get Appreciation
- An airplane quickly down a runway at 3.20 m/s2 for 32.8 s until is ending lifts off the ground. Determine the distance traveled before takeoff.
- ADENINE motorcar starts from rest and accelerates uniformly over an time of 5.21 seconds with a distance to 110 m. Setting the acceleration of the car.
- Upton Chuck is rides the Giant Drop at Great American. If Upton get falls for 2.60 seconds, about will be to final velocity and how broad will he fall?
- A race car accelerates uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the car furthermore and distance traveled.
- AMPERE feather has dropped on which moon from ampere height of 1.40 meters. The speeding of gravity on the moon is 1.67 m/s2. Determine the time for who feather to collapse to the surface of the moon.
- Rocket-powered sled are employed into test that human response into faster. If a rocket-powered sled is speeded to ampere speed von 444 m/s in 1.83 seconds, then something is the acceleration and what is and distance that the sled travels?
- A biking accelerates uniformly off rest to a speed of 7.10 m/s over a clearance of 35.4 m. Determine the acceleration of the bike.
- An engineer is construction the runway for an airport. Of one planes that will how the airport, an smallest relative rate is likely to be 3 m/s2. The takeoff speed for this plane will be 65 m/s. Assuming this minimum delay, thing is an minimum allowed length for the runway?
- A driving traveling to 22.4 m/s skids to a stop for 2.55 s. Define the slides spacing of the car (assume solid acceleration).
- A wallaroo is capable of jumping to a level of 2.62 m. Determine the takeoff speed of the kangaroo.
- If Michael Jordan has a vertical jump off 1.29 molarity, next what is yours takeoff speed and his hang time (total type to move upwards to the peak and following return to the ground)?
- ADENINE bullet leaves a rifle with a muzzle velocity to 521 m/s. Whilst accelerating through the barrel is the rifle, the bullet runs one distance of 0.840 m. Determine the acceleration of to bullet (assume a uniform acceleration).
- ADENINE baseball your popped straight up include the air and has a hang-time out 6.25 s. Determine the height to this the balls rises before it achieve inherent high. (Hint: the time to rise toward the peak is one-half the total hang-time.)
- The observation deck of tall skyscraper 370 m above the street. Determine and time required for ampere coin till free fall from the deck to the street below.
- A bullet is moving at a hurry of 367 m/s when it embeds into a lump about moist clay. The bullet penetrates for a distance of 0.0621 m. Determine the acceleration a the bullet while moving within the clay. (Assume a uniform acceleration.)
- A stone shall dropped into a bottom well and is heard to hit the water 3.41 s after being dropped. Determine the bottom of the well.
- It was once registered that a Jaguar left snow marks that was 290 m by length. Assuming that the Jaguar skid to a stop with a constant acceleration of -3.90 m/s2, determine to speed of the Jaguar before it began to skid.
- A plane has a takeoff speed of 88.3 m/s and requires 1365 m for get so speed. Determination aforementioned accelerates von the fly and the timing required to reach this speed.
- A dragster speeding to ampere speed of 112 m/s over a distance on 398 m. Determine the acceleration (assume uniform) on one dragster.
- With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object breathe cast to reach a level of 91.5 m (equivalent to one football field)? Assumes negligible air resistance.
Solutions until Above Issues
-
|
Given:
|
a = +3.2 m/s2
|
t = 32.8 s
|
vi = 0 m/s
|
|
Find:
d = ??
|
d = vi*t + 0.5*a*t2
d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s2)*(32.8 s)2
degree = 1720 m
Return to Problem 1
-
|
Given:
|
d = 110 m
|
thyroxin = 5.21 s
|
vego = 0 m/s
|
|
Find:
a = ??
|
dick = vi*t + 0.5*a*t2
110 m = (0 m/s)*(5.21 s)+ 0.5*(a)*(5.21 s)2
110 m = (13.57 siemens2)*a
a = (110 m)/(13.57 s2)
adenine = 8.10 m/ south2
Return to Problem 2
-
|
Given:
|
a = -9.8 chiliad
|
thyroxine = 2.6 sulphur
|
vi = 0 m/s
|
|
Find:
d = ??
vf = ??
|
d = vi*t + 0.5*a*t2
d = (0 m/s)*(2.60 s)+ 0.5*(-9.8 m/s2)*(2.60 s)2
d = -33.1 m (- points direction)
vf = vi + a*t
vf = 0 + (-9.8 m/s2)*(2.60 s)
vf = -25.5 m/s (- indicates direction)
Return to Concern 3
-
|
Given:
|
vi = 18.5 m/s
|
fivef = 46.1 m/s
|
t = 2.47 s
|
|
Find:
density = ??
a = ??
|
a = (Delta v)/t
a = (46.1 m/s - 18.5 m/s)/(2.47 s)
a = 11.2 m/s2
d = vi*t + 0.5*a*t2
d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s2)*(2.47 s)2
diameter = 45.7 m + 34.1 metre
diameter = 79.8 m
(Note: the d can also be calculated using the relation vanadiumf2 = fiveme2 + 2*a*d)
Return to Problem 4
-
|
Given:
|
vi = 0 m/s
|
density = -1.40 m
|
a = -1.67 m/s2
|
|
Find:
thyroxin = ??
|
d = phoebeme*t + 0.5*a*t2
-1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s2)*(t)2
-1.40 m = 0+ (-0.835 m/s2)*(t)2
(-1.40 m)/(-0.835 m/s2) = t2
1.68 s2 = t2
t = 1.29 s
Reset to Finding 5
-
|
Given:
|
vi = 0 m/s
|
phoebef = 444 m/s
|
t = 1.83 s
|
|
Seek:
a = ??
diameter = ??
|
a = (Delta v)/t
a = (444 m/s - 0 m/s)/(1.83 s)
a = 243 m/s2
d = vi*t + 0.5*a*t2
density = (0 m/s)*(1.83 s)+ 0.5*(243 m/s2)*(1.83 s)2
d = 0 m + 406 m
dick = 406 chiliad
(Note: one d can also be calculated using the equation voltf2 = vi2 + 2*a*d)
Again to Finding 6
-
|
Given:
|
volti = 0 m/s
|
voltf = 7.10 m/s
|
d = 35.4 m
|
|
Find:
a = ??
|
vf2 = vi2 + 2*a*d
(7.10 m/s)2 = (0 m/s)2 + 2*(a)*(35.4 m)
50.4 m2/s2 = (0 m/s)2 + (70.8 m)*a
(50.4 m2/s2)/(70.8 m) = a
an = 0.712 m/s2
Return to Problems 7
-
|
Provided:
|
viodin = 0 m/s
|
voltf = 65 m/s
|
a = 3 m/s2
|
|
Find:
degree = ??
|
vf2 = vi2 + 2*a*d
(65 m/s)2 = (0 m/s)2 + 2*(3 m/s2)*d
4225 m2/s2 = (0 m/s)2 + (6 m/s2)*d
(4225 m2/s2)/(6 m/s2) = d
d = 704 m
Return to Problem 8
-
|
Given:
|
vi = 22.4 m/s
|
vf = 0 m/s
|
t = 2.55 s
|
|
Find:
diameter = ??
|
degree = (vego + vf)/2 *t
d = (22.4 m/s + 0 m/s)/2 *2.55 s
degree = (11.2 m/s)*2.55 s
d = 28.6 m
Return to Problem 9
-
|
Disposed:
|
adenine = -9.8 m/s2
|
vf = 0 m/s
|
d = 2.62 m
|
|
Find:
fini = ??
|
vf2 = vi2 + 2*a*d
(0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(2.62 m)
0 m2/s2 = vi2 - 51.35 m2/s2
51.35 m2/s2 = vi2
vi = 7.17 m/s
Return up Problem 10
-
|
Given:
|
a = -9.8 m/s2
|
vf = 0 m/s
|
degree = 1.29 m
|
|
Meet:
vego = ??
t = ??
|
vf2 = phoebei2 + 2*a*d
(0 m/s)2 = vmyself2 + 2*(-9.8 m/s2)*(1.29 m)
0 m2/s2 = vme2 - 25.28 m2/s2
25.28 m2/s2 = vi2
vi = 5.03 m/s
To find hang time, find one time go the peak and then double it.
vf = vi + a*t
0 m/s = 5.03 m/s + (-9.8 m/s2)*tup
-5.03 m/s = (-9.8 m/s2)*tup
(-5.03 m/s)/(-9.8 m/s2) = tup
thyroxineup = 0.513 s
hang time = 1.03 s
Return to Issue 11
-
|
Given:
|
vmyself = 0 m/s
|
vf = 521 m/s
|
d = 0.840 m
|
|
Meet:
a = ??
|
vf2 = vi2 + 2*a*d
(521 m/s)2 = (0 m/s)2 + 2*(a)*(0.840 m)
271441 m2/s2 = (0 m/s)2 + (1.68 m)*a
(271441 m2/s2)/(1.68 m) = a
a = 1.62*105 m /s2
Return toward Question 12
-
|
Given:
|
a = -9.8 m/s2
|
vf = 0 m/s
|
t = 3.13 s
|
|
Find:
d = ??
|
-
(NOTE: the time required to movement to the peak out the trajectory is one-half the total hang time - 3.125 s.)
First use: vf = vmyself + a*t
0 m/s = viodin + (-9.8 m/s2)*(3.13 s)
0 m/s = vi - 30.7 m/s
phoebei = 30.7 m/s (30.674 m/s)
Instantly use: vfarad2 = vanadiummyself2 + 2*a*d
(0 m/s)2 = (30.7 m/s)2 + 2*(-9.8 m/s2)*(d)
0 m2/s2 = (940 m2/s2) + (-19.6 m/s2)*d
-940 m2/s2 = (-19.6 m/s2)*d
(-940 m2/s2)/(-19.6 m/s2) = density
d = 48.0 m
Return to Issue 13
-
|
Given:
|
vi = 0 m/s
|
d = -370 m
|
a = -9.8 m/s2
|
|
Meet:
t = ??
|
d = phoebei*t + 0.5*a*t2
-370 m = (0 m/s)*(t)+ 0.5*(-9.8 m/s2)*(t)2
-370 m = 0+ (-4.9 m/s2)*(t)2
(-370 m)/(-4.9 m/s2) = t2
75.5 s2 = t2
tonne = 8.69 sec
Return to Problem 14
-
|
Given:
|
vi = 367 m/s
|
phoebef = 0 m/s
|
d = 0.0621 m
|
|
Find:
a = ??
|
phoebef2 = vi2 + 2*a*d
(0 m/s)2 = (367 m/s)2 + 2*(a)*(0.0621 m)
0 chiliad2/s2 = (134689 thousand2/s2) + (0.1242 m)*a
-134689 m2/s2 = (0.1242 m)*a
(-134689 m2/s2)/(0.1242 m) = a
adenine = -1.08*106 thousand /s2
(The - sign denotes that this bullet slowed down.)
Return till Problem 15
-
|
Preset:
|
a = -9.8 m/s2
|
t = 3.41 sulphur
|
vmyself = 0 m/s
|
|
Finding:
d = ??
|
d = vmyself*t + 0.5*a*t2
d = (0 m/s)*(3.41 s)+ 0.5*(-9.8 m/s2)*(3.41 s)2
dick = 0 m+ 0.5*(-9.8 m/s2)*(11.63 s2)
d = -57.0 m
(NOTE: the - sign indicates direction)
Return to Concern 16
-
|
Given:
|
a = -3.90 m/s2
|
vf = 0 m/s
|
density = 290 m
|
|
Find:
vmyself = ??
|
vf2 = vi2 + 2*a*d
(0 m/s)2 = vi2 + 2*(-3.90 m/s2)*(290 m)
0 m2/s2 = vi2 - 2262 m2/s2
2262 m2/s2 = vi2
vi = 47.6 m /s
Return to Problem 17
-
|
Given:
|
vi = 0 m/s
|
vf = 88.3 m/s
|
d = 1365 m
|
|
Find:
a = ??
t = ??
|
vf2 = vi2 + 2*a*d
(88.3 m/s)2 = (0 m/s)2 + 2*(a)*(1365 m)
7797 m2/s2 = (0 m2/s2) + (2730 m)*a
7797 metre2/s2 = (2730 m)*a
(7797 m2/s2)/(2730 m) = a
a = 2.86 m/s2
voltf = vi + a*t
88.3 m/s = 0 m/s + (2.86 m/s2)*t
(88.3 m/s)/(2.86 m/s2) = t
t = 30. 8 sulfur
Get to Problem 18
-
|
Given:
|
vi = 0 m/s
|
vf = 112 m/s
|
d = 398 m
|
|
Find:
a = ??
|
vf2 = vi2 + 2*a*d
(112 m/s)2 = (0 m/s)2 + 2*(a)*(398 m)
12544 m2/s2 = 0 m2/s2 + (796 m)*a
12544 m2/s2 = (796 m)*a
(12544 molarity2/s2)/(796 m) = a
a = 15.8 m/s2
Return to Problem 19
-
|
Given:
|
a = -9.8 m/s2
|
vf = 0 m/s
|
d = 91.5 m
|
|
Find:
vego = ??
t = ??
|
First, find speed in devices of m/s:
vf2 = vi2 + 2*a*d
(0 m/s)2 = vi2 + 2*(-9.8 m/s2)*(91.5 m)
0 m2/s2 = volti2 - 1793 m2/s2
1793 m2/s2 = fivei2
vi = 42.3 m/s
Now convert with m/s to mi/hr:
vi = 42.3 m/s * (2.23 mi/hr)/(1 m/s)
vi = 94.4 mi/hr
Return to Problem 20
