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Lesson 7
Average Acceleration
While Turning
Done By Example
No Reference Pages
Uniform motion was previously defined as beingUniform motion was previously defined as being
constant velocity. And constant velocity requiresconstant velocity. And constant velocity requires
thatthat bothboth magnitudemagnitude andand directiondirection bebe constantconstant..
 If a car is moving at a constant 100 km/h aroundIf a car is moving at a constant 100 km/h around
a circular bend in the road, it will experiencea circular bend in the road, it will experience
acceleration because it is constantly changingacceleration because it is constantly changing
direction.direction.
 When aWhen aTurningTurning is calculated, it is in theis calculated, it is in the samesame
 direction asdirection as ΔΔvv and is anand is an averageaverage value (why?)value (why?)
 The equation used isThe equation used is aaTurningTurning == ΔΔv/v/ ΔtΔt
Sally is running south at 2.5 m/s. She then changes directionSally is running south at 2.5 m/s. She then changes direction
and runs east at the same pace. If it takes her 4 s to turn,and runs east at the same pace. If it takes her 4 s to turn,
determinedetermine aaTurningTurning
 G: vG: v11 = 2.5m/s [S], v= 2.5m/s [S], v22 ==
2.5 m/s [E],2.5 m/s [E], Δt = 4sΔt = 4s
 R:R: aaTurningTurning
 A:A: Δv = vΔv = v22 – v– v11 (draw vector(draw vector
diagram), then use the eq’ndiagram), then use the eq’n
for afor aturningturning
 Solution:Solution:
 Since the vSince the v22 and – vand – v11 areare ⊥⊥
to each other, and haveto each other, and have
the same magnitude,the same magnitude, thethe
angle θ is 45angle θ is 4500
 Statement:Statement:
∴∴ aaTurningTurning == 0.9 m/s2
[NE]
a T U R N I N G
= v 2
- v 1
t 2
- t 1
= 2 . 5 2
+ 2 . 5 2
[ N E ]
4
N
W E
v 2 = 2 . 5 m / s
-v1=2.5m/s
v
Practice Problems
Questions from McGraw-Hill TB
1.1. A car is traveling atA car is traveling at 50.0 km/h [N] when it makeswhen it makes
a turn which takesa turn which takes 5.0 s and then travels west atand then travels west at
the same pace. Det’m:the same pace. Det’m: Δv andand aTurning..
Ans. 71 km/h [SW], 3.9 m/s2
[SW],
2. A marble is rolling across a table with a velocity
of 21 cm/s [E]. The marble is tapped by a ruler
900
to its initial DOM. The marble accelerates in
the direction of the tap for 0.50 s at a rate of 56
cm/s2
. Det’m: v2
Ans. 35 cm/s [530
away from original direction]

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Lesson 7: Calculating Average Acceleration While Turning

  • 1. Lesson 7 Average Acceleration While Turning Done By Example No Reference Pages
  • 2. Uniform motion was previously defined as beingUniform motion was previously defined as being constant velocity. And constant velocity requiresconstant velocity. And constant velocity requires thatthat bothboth magnitudemagnitude andand directiondirection bebe constantconstant..  If a car is moving at a constant 100 km/h aroundIf a car is moving at a constant 100 km/h around a circular bend in the road, it will experiencea circular bend in the road, it will experience acceleration because it is constantly changingacceleration because it is constantly changing direction.direction.  When aWhen aTurningTurning is calculated, it is in theis calculated, it is in the samesame  direction asdirection as ΔΔvv and is anand is an averageaverage value (why?)value (why?)  The equation used isThe equation used is aaTurningTurning == ΔΔv/v/ ΔtΔt
  • 3. Sally is running south at 2.5 m/s. She then changes directionSally is running south at 2.5 m/s. She then changes direction and runs east at the same pace. If it takes her 4 s to turn,and runs east at the same pace. If it takes her 4 s to turn, determinedetermine aaTurningTurning  G: vG: v11 = 2.5m/s [S], v= 2.5m/s [S], v22 == 2.5 m/s [E],2.5 m/s [E], Δt = 4sΔt = 4s  R:R: aaTurningTurning  A:A: Δv = vΔv = v22 – v– v11 (draw vector(draw vector diagram), then use the eq’ndiagram), then use the eq’n for afor aturningturning  Solution:Solution:  Since the vSince the v22 and – vand – v11 areare ⊥⊥ to each other, and haveto each other, and have the same magnitude,the same magnitude, thethe angle θ is 45angle θ is 4500  Statement:Statement: ∴∴ aaTurningTurning == 0.9 m/s2 [NE] a T U R N I N G = v 2 - v 1 t 2 - t 1 = 2 . 5 2 + 2 . 5 2 [ N E ] 4 N W E v 2 = 2 . 5 m / s -v1=2.5m/s v
  • 4. Practice Problems Questions from McGraw-Hill TB 1.1. A car is traveling atA car is traveling at 50.0 km/h [N] when it makeswhen it makes a turn which takesa turn which takes 5.0 s and then travels west atand then travels west at the same pace. Det’m:the same pace. Det’m: Δv andand aTurning.. Ans. 71 km/h [SW], 3.9 m/s2 [SW], 2. A marble is rolling across a table with a velocity of 21 cm/s [E]. The marble is tapped by a ruler 900 to its initial DOM. The marble accelerates in the direction of the tap for 0.50 s at a rate of 56 cm/s2 . Det’m: v2 Ans. 35 cm/s [530 away from original direction]