NAME__** Answers**_____ PHYSICS
ACC PROBLEMS 1

Solve with: formulas, D-T and V-T graphs, or Interactive Physics!

D= Vavg T Vavg = (Vi + Vf) / 2 (if A is constant) A = (Vf-Vi) /T

D-T graph, slope is velocity V-T graph area is displacement, slope is accel

1) While driving his sports car at 20 m/s Eddie comes up behind a slow moving dump truck and decides to pass it. If he can accelerate at a rate of 5 m/s/s, how long will it take him to reach a speed of 30 m/s ?

*Vi= 20 m/s, A=
5 m/s^2, Vf= 30 m/s, T = ??*

* *

*A = (Vf-Vi)/T 5
= (30-20)/T 5T
= 10 T
= 2 sec*

2) A jet taxing down a runway receives word that it must return to the hangar for an important message. The jet is traveling at 45 m/s . What is the acceleration of the plane if it takes 5 sec to bring it to a halt?

*Vi = 45 m/s, Vf=
0, T
= 5 sec, A=
??*

* *

*A= (Vf-Vi)/T A=
(45-0)/5 A= 9
m/s^2*

3) Hans stands at the rim of the Grand Canyon and yodels. He hears his yodel 5 sec later. (2.5 sec to go down, 2.5 sec to come back) Assume that the speed of sound is 340 m/s . How deep is the canyon?

*T= 5/2 = 2.5 sec, Vavg=
340 m/s D=??*

* *

*D=VavgT D=
340*2.5 = 840 m*

4) Monica is walking to the hairdresser at 1.3 m/s, then realizes she is going to be late. She quickens her pace at the rate of .09 m/s/s . What is her speed after 10 s? How far has she traveled in this time?

* Vi = 1.3
m/s, A = .09 m/s^2, T= 10 sec, Vf=??, D=??*

* *

*A= (Vf-Vi)/T .09
= (Vf-1.3)/10 10*.09
= Vf-1.3 *

*Vf=1.3+.9 =
2.2 m/s=Vf*

* *

*D=VavgT Vavg=(Vi+Vf)/2 Vavg=(1.3+2.2)/2 = 1.75 m/s=Vavg*

* *

*D = (1.75 m/s) * 10 sec = 17.5 m=D*

5) Bobby wants to catch up with his friend in the hallway. His friend is 10 m ahead of him, walking at 3 m/s. If Bobby constantly accelerates at a rate of 2 m/s/s from rest, how long till he catches up to his friend? How far? What is his speed at that time?

*Friend: Vavg=
3m/s D=10+VavgT D=
10 +3T*

*Bobby: Vi=0 Vavg=(Vi+Vf)/2=Vf/2 A=
2 m/s^2*

* *

* A=(Vf-Vi)/T 2
= Vf/T Vf=2T*

* D=
Vavg T D=
(Vf/2)T D=
(2T/2)T D=T^2*

* *

* Friend D=10+3T Bobby D=T^2*

* 10+3T = T^2*

* 0
= T^2-3T-10*

* Solve…..
T=+5 or -2, so obviously T=+5 seconds!*

* D=10+3T=10+3(5)=
25 m or D =T^2=5^2=25m*

* Bobby
Vf=0+2(T)=2(5)=10 m/s*

* *

*Better to solve this one on a graph!!!!*

* *

*On a V-T graph, look to where the areas under the
graph are the same…. Not where the velocities are equal!*

* *

*See car passing van interactive physics for an example
of this!*

** HONORS

6) Frank is going 4 m/s for 5 seconds. If I start 2 m ahead of him going 9 m/s, at what rate will I have to deaccelerate so I will be in the same spot as him after 5 sec?

*Frank: Vavg= 4m/s, T= 5 sec, D= VavgT D=4(5)=20 m*

* *

*me: Di=2, Vi=9m/s, T=5sec, Vf=?, A=?, Df=20 m*

* D=
20-2=18m = Vavg T=Vavg (5)*

* Vavg
= D/T = 18/5 = 3.6 m/s*

* Vavg
=( Vi+Vf)/2 3.6
= (9+Vf)/2*

* 7.2
=9 + Vf Vf= -1.8
m/s*

* A=
(Vf-Vi)/T A=
( -1.8-9)/5 = -10.8/5 = -2.16 m/s^2
=A*

7) A bird is flying along at 60 m/s and reaches a constantly increasing force of wind which causes the bird to deaccelerate at the rate of 12 m/s/s. How far will the bird have gone after 15 seconds? What will its velocity be at that time?

*Vi = 60 m/s, A = -12 m/s^2, T= 15 s, D =??, Vf =??*

* *

*A= (Vf-Vi)/T -12
= (Vf-60)/15 Vf=
60 + -12(15) = -120 m/s= Vf*

*D=VavgT D
= (Vi+Vf)/2 T D=
(60+-120)/2 (15) = -30*15 = -450 m = D*

8) What speed do I have to start at if I want to go 200 meters accelerating at the rate of 4 m/s/s for 10 seconds?

*Vi = ??, D= 200m, A = 4 m/s^2, T = 10 s*

* *

*D= (Vavg)T 200=Vavg(10) Vavg
= 200/10 =20
m/s*

* *

*A= (Vf-Vi)/T 4
= (Vf-Vi)/10 40
= Vf-Vi Vf=40+Vi*

*Vavg = (Vi+Vf)/2 20
= (Vi+Vf)2 40=Vi+Vf 40=Vi+40+Vi *

*0=2Vi,
Vi=0!!*