122

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12–114.

The automobile has a speed of 80 ft>s at point A and an acceleration a having a magnitude of 10 ft>s2, acting in the direction shown. Determine the radius of curvature of the path at point A and the tangential component of acceleration.

SOLUTIONAcceleration: The tangential acceleration is

at = a cos 30° = 10 cos 30° = 8.66 ft>s2 Ans.

and the normal acceleration is an = a sin 30° = 10 sin 30° = 5.00 ft>s2. Applying

Eq. 12–20, an =v2

r, we have

r =v2

an=

802

5.00= 1280 ft Ans.

Ans:at = 8.66 ft>s2

r = 1280 ft

123

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12–115.

The automobile is originally at rest at If its speed isincreased by where t is in seconds,determine the magnitudes of its velocity and accelerationwhen t = 18 s.

v# = 10.05t22 ft>s2,

s = 0.

SOLUTION

When

Therefore the car is on a curved path.

Ans.

Ans.a = 42.6 ft>s2

a = 2(39.37)2 + (16.2)2

at = 0.05(182) = 16.2 ft/s2

an =(97.2)2

240= 39.37 ft>s2

v = 0.0167(183) = 97.2 ft>s

t = 18 s, s = 437.4 ft

s = 4.167(10- 3) t4

L

s

0ds =

L

t

00.0167 t3 dt

v = 0.0167 t3

L

v

0dv =

L

t

00.05 t2 dt

at = 0.05t2

s

240 ft

300 ft

Ans:v = 97.2 ft>sa = 42.6 ft>s2

124

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*12–116.

SOLUTIONThe car is on the curved path.

So that

Ans.

Ans.a = 2(55.51)2 + (18.17)2 = 58.4 ft>s2

at = 0.05(19.06)2 = 18.17 ft>s2

an =(115.4)2

240= 55.51 ft>s2

v = 115 ft>s

v = 0.0167(19.06)3 = 115.4

t = 19.06 s

550 = 4.167(10-3) t4

s = 4.167(10- 3) t4

L

s

0ds =

L

t

00.0167 t3 dt

v = 0.0167 t3

L

v

0dv =

L

t

00.05 t2 dt

at = 0.05 t2

The automobile is originally at rest If it then starts toincrease its speed at where t is in seconds,determine the magnitudes of its velocity and acceleration ats = 550 ft.

v# = 10.05t22 ft>s2,

s = 0.s

240 ft

300 ft

Ans:v = 115 ft>sa = 58.4 ft>s2

128

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

*12–120.

The car travels along the circular path such that its speed isincreased by , where t is in seconds.Determine the magnitudes of its velocity and accelerationafter the car has traveled starting from rest.Neglect the size of the car.

s = 18 m

at = (0.5et) m>s2

SOLUTION

Solving,

Ans.

Ans.a = 2a2t + a2

n = 220.352 + 13.142 = 24.2 m>s2

an =v2

r=

19.852

30= 13.14 m>s2

at = v# = 0.5et ƒ t = 3.7064 s = 20.35 m>s2

v = 0.5(e3.7064 - 1) = 19.85 m>s = 19.9 m>s

t = 3.7064 s

18 = 0.5(et - t - 1)

L

18

0ds = 0.5

L

t

0(e t - 1)dt

v = 0.5(e t - 1)

L

v

0dv =

L

t

00.5e t dt

s 18 m

30 mρ

Ans:v = 19.9 m>sa = 24.2 m>s2

129

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12–121.

SOLUTIONVelocity:The speed of the car at B is

Radius of Curvature:

Acceleration:

When the car is at

Thus, the magnitude of the car’s acceleration at B is

Ans.a = 2at2 + an

2 = 2(-2.591)2 + 0.91942 = 2.75 m>s2

a t = C0.225 A51.5 B - 3.75 D = -2.591 m>s2

B As = 51.5 m B

a t = vdv

ds= A25 - 0.15s B A -0.15 B = A0.225s - 3.75 B m>s2

an =vB

2

r=

17.282

324.58= 0.9194 m>s2

r =B1 + a

dy

dxb

2 B3>2

2 d2y

dx22

=c1 + a -3.2 A10-3 Bxb

2

d3>2

2 -3.2 A10-3 B 24

x = 50 m

= 324.58 m

d2y

dx2 = -3.2 A10-3 B

dy

dx= -3.2 A10-3 Bx

y = 16 -1

625x2

vB = C25 - 0.15 A51.5 B D = 17.28 m>s

The car passes point A with a speed of after which itsspeed is defined by . Determine themagnitude of the car’s acceleration when it reaches point B,where and x = 50 m.s = 51.5 m

v = (25 - 0.15s) m>s25 m>s

y 16 x21625

y

s

x16 m

B

A

Ans:a = 2.75 m>s2

180

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12–171.

At the instant shown, the man is twirling a hose over his head with an angular velocity u

# = 2 rad>s and an angular

acceleration u$ = 3 rad>s2. If it is assumed that the hose lies

in a horizontal plane, and water is flowing through it at a constant rate of 3 m>s, determine the magnitudes of the velocity and acceleration of a water particle as it exits the open end, r = 1.5 m.

� 2 rad/s ·u

� 3 rad/s2 · ·u

u

r � 1.5 m

Solutionr = 1.5

r# = 3

r$ = 0

u#= 2

u$= 3

vr = r# = 3

vu = ru#= 1.5(2) = 3

v = 2(3)2 + (3)2 = 4.24 m>s Ans.

ar = r$ - r(u

#)2 = 0 - 1.5(2)2 = 6

au = r u$

+ 2r#u#= 1.5(3) + 2(3)(2) = 16.5

a = 2(6)2 + (16.5)2 = 17.6 m>s2 Ans.

Ans:v = 4.24 m>sa = 17.6 m>s2

184

© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

12–175.

SOLUTION

Ans.

Ans.a = r$ - ru

#2 + ru

$+ 2 r

#u#

2 = [4 - 2(6)2 + [0 + 2(4)(6)]2

= 83.2 m s2

v = 3Ar# B + Aru#B2 = 2 (4)2 + [2(6)]2 = 12.6 m>s

r = 2t2 D10 = 2 m

L

1

0dr =

L

1

04t dt

u = 6 u$

= 0

r = 4t|t = 1 = 4 r# = 4

A block moves outward along the slot in the platform witha speed of where t is in seconds. The platformrotates at a constant rate of 6 rad/s. If the block starts fromrest at the center, determine the magnitudes of its velocityand acceleration when t = 1 s.

r# = 14t2 m>s,

θ

θ· = 6 rad/sr

$

#

2

2 ]2

Ans:

v = 12.6 m>sa = 83.2 m>s2