Practice 2E: | 1 | 2 | 3 | 4 | 5 | 6 | Go up
Final velocity after any displacement - by Kevin Bailey & Chicken Breast, 2002

1. Find the velocity after the stroller has traveled 6.32 m.

<solution goes here>

2. A car traveling initially at +7.0 m/s accelerates uniformly at the rate of +0.80 m/s2 for a distance of 245 m.

a.  What is its velocity at the end of the acceleration?

b.  What is its velocity after it accelerates for 125 m?

c.  What is its velocity after it accelerates for 67 m?

Ok, so first to solve for the Final veloctiy.  You need to know this formula.  vf2 = vi2 + 2a(Dx), where:

Dx = 245 m

vi = +7.0 m/s

Dt = ???

vf = ???

a = +0.80 m/s2

So, vf2 = (7.0 m/s)2 + 2(0.80 m/s2)(245 m)  = 735 m/s, Final Velocity = 735(1/2) = 27.1 m/s

Got it?  Good.  Alrighty then, to solve for the speed this is all you need:  vf2 = vi2 + 2a(Dx), vf = ?, vi = +7.0 m/s, a = 0.80 m/s2, Dx = 125 m

So that's vf2 = (7.0 m/s)2 + 2(0.80 m/s2)(125 m) = 249 m/s, Final Velocity = 249(1/2) = 15.8 m/s

Yes, and finally for part c.  So then, to solve for the speed this is all you need:  vf2 = vi2 + 2a(Dx), vf = ?, vi = +7.0 m/s, a = 0.80 m/s2, Dx = 67 m

So that's vf2 = (7.0 m/s)2 + 2(0.80 m/s2)(67 m) = 156.2 m/s, Final Velocity = 249(1/2) = 12.5 m/s

3.A car accelerates uniformly in a straight line from rest at the rate of 2.3 m/s2.

a.  What is the speed of the car after it has traveled 55 m?

b.  How long des it take the car to travel 55 m?

Ok, so first to solve for the Final velocity.  You need to know this formula.  vf2 = vi2 + 2a(Dx), where:

Dx = 55 m

vi = 0 m/s

Dt = ???

vf = ???

a = 2.3 m/s2

So, vf2 = (0 m/s)2 + 2(2.3 m/s2)(55 m)  = 253 m/s, Final Velocity = 253(1/2) = 15.9 m/s

Ok, second to solve for the time.  You need to know this formula.  Dx = 1/2(vi + vf)Dt, where:

Dx = 55 m

vi = 0 m/s

Dt = ???

vf = 15.9 m/s

a = 2.3 m/s2

So, 55 m = 1/2(0 + 15.9 m/s)Dt so Dt = (55 m) / (7.95 m/s) = 6.9 s

4. A certain car is capable of accelerating at a uniform rate of 0.85 m/s2.  What is the magnitude of the car's displacement as it accelerates uniformly from a speed of 83 km/h to one of 94 km/h?

Humm...this is a tricky one.  In order to find the displacement you need to know this formula.  vf2 = vi2 + 2a(Dx), where:

Dx = ???

vi = 83 km/h or 23 m/s

Dt = ???

vf = 94 km/h or 26 m/s

a = 0.85 m/s2

So just plug in the numbers and let that magical math machine do the rest!

(26 m/s)2 = (23 m/s)2 + 2(0.85 m/s2)(Dx)  so Dx = 86.5 m

5. An aircraft has a liftoff speed of 120 km/h.  What minimum uniform acceleration does this require if the aircraft is to be airborne after a takeoff run of 240 m?

Well now what could that acceleration be?  In order to find the acceleration you need to know this formula.  vf2 = vi2 + 2a(Dx), where:

Dx = 240 m

vi = 0 m/s

Dt = ???

vf = 120 km/h or 33.3 m/s (OH, km/h to m/s is to multiply by 1000/3600)

a = ??? m/s2

Uh oh, I hear that familiar chomping sound of the Math Robot coming again!  Don't worry, just plug and chicken breast it.

(33.3 m/s)2 = (0 m/s)2 + 2(??? m/s2)(240 m)  so a = 2.31 m/s2

6. A motorboat accelerates uniformly from a velocity of 6.5 m/s to the west to a velocity of 1.5 m/s to the west.  If its acceleration was 2.7 m/s2 to the east, how far did it travel during the acceleration?

Holy rusted metal Batman!  In order to find the displacement you need to know this formula.  vf2 = vi2 + 2a(Dx), where:

Dx = ???

vi = 6.5 m/s

Dt = ???

vf =1.5 m/s

a = -2.7 m/s2

"Ok Spock, we're almost," said Capt. Kirk.

(1.5 m/s)2 = (6.5 m/s)2 + 2(-2.7 m/s2)(Dx)  so Dx= 7.4 m

SONIC BOOM!!!