Answer:
(A) Original speed= 9.22 m/s
(B) Acceleration= -1.0099 m\s^2
Explanation:
A truck covers 40m in 7.10 secs
The truck slows down at a uniform velocity of 2.05 m/s
(A) The original speed can be calculated as follows
Vo= 2(40)/7.10 - 2.05
= 80/7.10 - 2.05
= 11.2676 - 2.05
= 9.22m/s
(B) The acceleration can be calculated as follows
a= Vf-Vo/t
= 2.05-9.22/7.10
= -7.17/7.10
= -1.0099m/s^2
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Answer:
Explanation:
Time taken to accelerate to 28 m /s
= 28 / 2 = 14 s
a ) Total length of time in motion
= 14 + 41 + 5
= 60 s .
b )
Distance covered while accelerating
s = ut + 1/2 at²
= 0 + .5 x 2 x 14²
= 196 m .
Distance covered while moving in uniform motion
= 28 x 41
= 1148 m
distance covered while decelerating
v = u - at
0 = 28 - a x 5
a = 5.6 m / s²
v² = u² - 2 a s
0 = 28² - 2 x 5.6 x s
s = 28² / 2 x 5.6
= 70 m .
Total distance covered
= 196 + 1148 + 70
= 1414 m
total time taken = 60 s
average velocity
= 1414 / 60
= 23.56 m /s .
Answer:
10
Explanation:
i = 5/.5 = 10 Amps. Hope this helps :)