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A record of travel along a straight path is as follows: (a) Start from rest with constant acceleration of 3.9 m/s 2 for 15.3 s; (b) Constant
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A record of travel along a straight path is as follows: (a) Start from rest with constant acceleration of 3.9 m/s 2 for 15.3 s; (b) Constant velocity of 59.67 m/s for the next 0.934 min; (c) Constant negative acceleration of −11.1 m/s 2 for 3.71 s. What was the total displacement
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3 years
2021-08-06T12:01:53+00:00
2021-08-06T12:01:53+00:00 2 Answers
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Answers ( )
Answer:
a) 456.48m
b) 3343.91m
c) – 76.376m
Total displacement = 3724.01m
Explanation:
Using kinematic equation for displacement
◇x1 = Vo t + 1/2 at^2
Where ◇x1 = displacement
Vo = initial velocity
a = acceleration
t = time
◇x1 = 0(15.3) + 1/2 (3.9)(15.3)^2
◇x1 = 0 + 456.48
Displacement = 456.48m
b) ◇x2 = Vt1
V = velocity = 59.67m/s
◇x2 = 59.67 ×0.934minute×60seconds
◇x2 = 59.67 ×56.04 = 3343.91
c)
a= -11.1m/s
= (-11.1 m/s^2× 3.71seconds)
◇ x3 =1/2 (41.18m/s ×3.71)
◇x3 =- 76.376m
Total displacement = ◇x1 + ◇x2 + ◇x3
Total displacement = 456.48 + 3343.91 – 76.376
Total displacement = 3724.01m
Answer:
Total displacement Is 1410.9945 meters
Explanation:
The total displacement is as follows.
The first part = triangle A
3.9 m/s² for 15.3 sec.
3.9 * 15.3 = 59.67 m/s
Second part = rectangle
15.67 m/s for (0.934*60)sec= 56.04 sec
Third part = triangle B
11.1m/s² for 3.71 sec
11.1 * 3.71 = 41.171 m/s
The area if these shapes gives the displacement.
For A = 1/2(15.3*59.67)
= 456.4755 m
For B = 15.67 * 56.04
= 878.1468 m
For C = 1/2(41.171*3.71)
= 76.372
Total displacement = 456.4755 + 878.1468 + 76.3722
= 1410.9945 meters