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**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/s ^{2} 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.
v_{f}^{2} = v_{i}^{2} + 2a(Dx),
where:

Dx = 245 m

v_{i} = +7.0 m/s

Dt = ???

v_{f} = ???

a = +0.80 m/s^{2}

So, v_{f}^{2} = (7.0 m/s)^{2} + 2(0.80 m/s^{2})(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: v_{f}^{2} = v_{i}^{2} + 2a(Dx),
v_{f} = ?, v_{i} = +7.0 m/s, a = 0.80 m/s^{2}, Dx
= 125 m

So that's v_{f}^{2} = (7.0 m/s)^{2} + 2(0.80 m/s^{2})(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: v

_{f}^{2}= v_{i}^{2}+ 2a(Dx), v_{f}= ?, v_{i}= +7.0 m/s, a = 0.80 m/s^{2}, Dx = 67 m

So that's v_{f}^{2} = (7.0 m/s)^{2} + 2(0.80 m/s^{2})(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/s ^{2}.**

** 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. v

_{f}^{2}= v_{i}^{2}+ 2a(Dx), where:Dx = 55 m

v

_{i}= 0 m/sDt = ???

v

_{f}= ???a = 2.3 m/s

^{2}So, v

_{f}^{2}= (0 m/s)^{2}+ 2(2.3 m/s^{2})(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}(v_{i}+ v_{f})Dt, where:Dx = 55 m

v

_{i}= 0 m/sDt = ???

v

_{f}= 15.9 m/sa = 2.3 m/s

^{2}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/s ^{2}. 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. v

_{f}^{2}= v_{i}^{2}+ 2a(Dx), where:Dx = ???

v

_{i}= 83 km/h or 23 m/sDt = ???

v

_{f}= 94 km/h or 26 m/sa = 0.85 m/s

^{2}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/s^{2})(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. v

_{f}^{2}= v_{i}^{2}+ 2a(Dx), where:Dx = 240 m

v

_{i}= 0 m/sDt = ???

v

_{f}= 120 km/h or 33.3 m/s (OH, km/h to m/s is to multiply by 1000/3600)a = ??? m/s

^{2}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/s^{2})(240 m) so a =2.31 m/s^{2}

**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/s ^{2} 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. v

_{f}^{2}= v_{i}^{2}+ 2a(Dx), where:Dx = ???

v

_{i}= 6.5 m/sDt = ???

v

_{f}=1.5 m/sa = -2.7 m/s

^{2}"Ok Spock, we're almost," said Capt. Kirk.

(1.5 m/s)

^{2}= (6.5 m/s)^{2}+ 2(-2.7 m/s^{2})(Dx) so Dx=7.4 m

SONIC BOOM!!!

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