1. An object's position as a function of time is for it first 4 seconds of motion is given by the expression beginning at t = 0; x(t) + 3t2 + 2e0.5t (Assume constants have proper SI Units) (t+1)² a) What is the object's average velocity over the first 4.00 seconds? -0.510) X (0) = 3(0) (0+1)2 %3D -0.5(4) x(4)= + 3(4) + 2 e (4+1)? + 98 + 0. 271-48.5 %3D 48.5-7 =10.4m/s b) What is the object's instantaneous velocity at 2.00 seconds into its motion? 2. 5(t +1) - 0.5t +36? + 2e x(t) = -10 (t+5 + 6t -0.5t -e 10 X (2) = +6(2)--0.5(2) %3D (2+1) + 12 - e 27 - 0.3764 + 12 - 0.3679 =\||.3 m/s c) What is the object's average acceleration over the first 4.00 seconds? x'(t)= + 6t - O.St (t+1)3 x'(4) + 6(4)- e - O. 5(4) x'C0) = + 6(0) - e -0.5(0) %3D (4+1)3 lo+1)3 -2 + 24 -10 +0 - | 125 -0.08+24 - 0.13 53 -11 %3D = 23.8 23.8 - (-11) 8.7m/s< %3D d) At what time is the object's velocity 10.0 m/s? Set up expression for partial credit. 2. X(t) = 10.= + 3t + 2e (t+1) t=1.675 This is Postion -05 (1.67) -0.5 + 3 + 2e 21 5 0.1013 + 8.3667+ 0.8677 =9.9 -as(1.68) 725+3 + 1.213 = 5.43 %3D 7 =(1.67+)2+ 3(1.679+ 2e as + 3(2)+ 2 eoas 0.6961+ 8.4672 t 0.8634=10,02 0.(1,69) 311.69)?20(1,69) 2+1 2+3(1.68)+2e (1.6811)" + 3(4) + 0.73 58313.29 69= (1.69+)² 311.5) + 2e -0.S (. is +1)2 +3(1,5)? + 2ē (1.s +1)2 0.6409 + 85683 +6.859 1=10.1| (טרו)*0:. ז "(0 ר .)3 0.8+ 6.75 0.9447 = Q: + 0.9447= ,70= (1.70+1) -0.5(1.70) 8- 1.60- + 3(1.60) + 2 e - 0.S(1.6c 0.68587 t 861 t08548 (1.60 +1) =16.21 1.923+7.68 108987= 0.7796 + 7.6g to8987=a
1. An object's position as a function of time is for it first 4 seconds of motion is given by the expression beginning at t = 0; x(t) + 3t2 + 2e0.5t (Assume constants have proper SI Units) (t+1)² a) What is the object's average velocity over the first 4.00 seconds? -0.510) X (0) = 3(0) (0+1)2 %3D -0.5(4) x(4)= + 3(4) + 2 e (4+1)? + 98 + 0. 271-48.5 %3D 48.5-7 =10.4m/s b) What is the object's instantaneous velocity at 2.00 seconds into its motion? 2. 5(t +1) - 0.5t +36? + 2e x(t) = -10 (t+5 + 6t -0.5t -e 10 X (2) = +6(2)--0.5(2) %3D (2+1) + 12 - e 27 - 0.3764 + 12 - 0.3679 =\||.3 m/s c) What is the object's average acceleration over the first 4.00 seconds? x'(t)= + 6t - O.St (t+1)3 x'(4) + 6(4)- e - O. 5(4) x'C0) = + 6(0) - e -0.5(0) %3D (4+1)3 lo+1)3 -2 + 24 -10 +0 - | 125 -0.08+24 - 0.13 53 -11 %3D = 23.8 23.8 - (-11) 8.7m/s< %3D d) At what time is the object's velocity 10.0 m/s? Set up expression for partial credit. 2. X(t) = 10.= + 3t + 2e (t+1) t=1.675 This is Postion -05 (1.67) -0.5 + 3 + 2e 21 5 0.1013 + 8.3667+ 0.8677 =9.9 -as(1.68) 725+3 + 1.213 = 5.43 %3D 7 =(1.67+)2+ 3(1.679+ 2e as + 3(2)+ 2 eoas 0.6961+ 8.4672 t 0.8634=10,02 0.(1,69) 311.69)?20(1,69) 2+1 2+3(1.68)+2e (1.6811)" + 3(4) + 0.73 58313.29 69= (1.69+)² 311.5) + 2e -0.S (. is +1)2 +3(1,5)? + 2ē (1.s +1)2 0.6409 + 85683 +6.859 1=10.1| (טרו)*0:. ז "(0 ר .)3 0.8+ 6.75 0.9447 = Q: + 0.9447= ,70= (1.70+1) -0.5(1.70) 8- 1.60- + 3(1.60) + 2 e - 0.S(1.6c 0.68587 t 861 t08548 (1.60 +1) =16.21 1.923+7.68 108987= 0.7796 + 7.6g to8987=a
College Physics
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Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Answer Part D). At what time is the object's velocity 10 m/s?
Transcribed Image Text:1. An object's position as a function of time is for it first 4 seconds of motion is
given by the expression beginning at t = 0;
x(t)
+ 3t2 + 2e0.5t (Assume constants have proper SI Units)
(t+1)²
a) What is the object's average velocity over the first 4.00 seconds?
-0.510)
X (0) =
3(0)
(0+1)2
%3D
-0.5(4)
x(4)=
+ 3(4) + 2 e
(4+1)?
+ 98 + 0. 271-48.5
%3D
48.5-7
=10.4m/s
b) What is the object's instantaneous velocity at 2.00 seconds into its motion?
2.
5(t +1)
- 0.5t
+36?
+ 2e
x(t) = -10 (t+5 + 6t
-0.5t
-e
10
X (2) =
+6(2)--0.5(2)
%3D
(2+1)
+ 12 - e
27
- 0.3764 + 12 - 0.3679 =\||.3 m/s
Transcribed Image Text:c) What is the object's average acceleration over the first 4.00 seconds?
x'(t)=
+ 6t
- O.St
(t+1)3
x'(4)
+ 6(4)- e
- O. 5(4)
x'C0) =
+ 6(0) - e
-0.5(0)
%3D
(4+1)3
lo+1)3
-2
+ 24
-10 +0 - |
125
-0.08+24
- 0.13 53
-11
%3D
= 23.8
23.8 - (-11)
8.7m/s<
%3D
d) At what time is the object's velocity 10.0 m/s? Set up expression for partial
credit.
2.
X(t) = 10.=
+ 3t + 2e
(t+1)
t=1.675
This is Postion
-05 (1.67)
-0.5
+ 3 + 2e
21
5 0.1013 + 8.3667+ 0.8677 =9.9
-as(1.68)
725+3 + 1.213 = 5.43
%3D
7 =(1.67+)2+ 3(1.679+ 2e
as
+ 3(2)+ 2 eoas
0.6961+ 8.4672 t 0.8634=10,02
0.(1,69)
311.69)?20(1,69)
2+1
2+3(1.68)+2e
(1.6811)"
+ 3(4) + 0.73 58313.29
69=
(1.69+)²
311.5) + 2e
-0.S (.
is +1)2 +3(1,5)? + 2ē
(1.s +1)2
0.6409 + 85683 +6.859 1=10.1|
(טרו)*0:. ז "(0 ר .)3
0.8+ 6.75 0.9447 = Q:
+ 0.9447=
,70=
(1.70+1)
-0.5(1.70)
8-
1.60-
+ 3(1.60) + 2 e
- 0.S(1.6c
0.68587 t 861 t08548
(1.60 +1)
=16.21
1.923+7.68 108987=
0.7796 + 7.6g to8987=a
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