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A particle starts from rest and accelerates as shown in the figure below.(m/s2)a1t (s)15 20510-1-2(a) Determine the particle's speed at t = 10.0 s and at t = 20.0 s.10.0 sm/stt 20.0 sm/s(b) Determine the distance traveled in the first 20.0 s. (Enter your answer to one decimal place.)m

Question
A particle starts from rest and accelerates as shown in the figure below.
(m/s2)
a
1
t (s)
15 20
5
10
-1
-2
(a) Determine the particle's speed at t = 10.0 s and at t = 20.0 s.
10.0 s
m/s
t
t 20.0 s
m/s
(b) Determine the distance traveled in the first 20.0 s. (Enter your answer to one decimal place.)
m
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A particle starts from rest and accelerates as shown in the figure below. (m/s2) a 1 t (s) 15 20 5 10 -1 -2 (a) Determine the particle's speed at t = 10.0 s and at t = 20.0 s. 10.0 s m/s t t 20.0 s m/s (b) Determine the distance traveled in the first 20.0 s. (Enter your answer to one decimal place.) m

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Step 1

Between t = 0 s and t = 10 s, the particle’s acceleration is 2 m/s2.

Use kinematics’ first equation to find the velocity v10 of the particle at t =10.0 s as,

10
10
(0 m/s)+(2 m/s(10 s}-{0 s})
- 20 m/s
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10 10 (0 m/s)+(2 m/s(10 s}-{0 s}) - 20 m/s

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Step 2

Here, v0 is the particle’s velocity at t = 0 s and a1 is the particle’s acceleration between 0 s and 10 s.

Between t = 10 s and t = 15 s, the particle’s acceleration is 0 m/s2.

Use kinematics’ first equation to find the particle’s velocity v15 at t =15.0 s as,

= V.
10
15
10
=(20 m/s)+(0 m/s(15 s-10 s)
= 20 m/s
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= V. 10 15 10 =(20 m/s)+(0 m/s(15 s-10 s) = 20 m/s

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Step 3

Here, a2 is the particle’s acceleration between 10 s and 15 s.

 

Between t = 15 s and t = 20 s, the particle’s acceleration is -3 m/s2.

Use kinema...

V20 is +a (20-4s)
(20 m/s)+(-3 m/s) ({20 s}-{15 s})
=(20 m/s)-(15 m/s)
15
15
= 5 m/s
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V20 is +a (20-4s) (20 m/s)+(-3 m/s) ({20 s}-{15 s}) =(20 m/s)-(15 m/s) 15 15 = 5 m/s

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Physics

Kinematics

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