Problem:
A jeepney from rest accelerates uniformly over a time of 3.25 second and covers a distance of 15m. Determined the acceleration of the jeepney.
Given Data:
d (distance) = 15 m
Vi (initial velocity) = 0 m/s (from rest)
t (time) = 3.25 s
a (acceleration) = unknown
Solution:
We can use the formula (d = Vit + 1/2 at² ) and to find a (acceleration) we can use the formula a = d / 1/2 (t²) since Vi = 0 m/s.
a = d / 1/2 (t²)
= 15 m / 1/2 (3.25 s)²
= 15 m / 1/2 (10.56 s²)
= 15 m/ 5.28 s²
= 2.84 m/s²
Answer:
the acceleration of the jeepney is 2.84 m/s²
Explanation:
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Answers & Comments
Problem:
A jeepney from rest accelerates uniformly over a time of 3.25 second and covers a distance of 15m. Determined the acceleration of the jeepney.
Given Data:
d (distance) = 15 m
Vi (initial velocity) = 0 m/s (from rest)
t (time) = 3.25 s
a (acceleration) = unknown
Solution:
We can use the formula (d = Vit + 1/2 at² ) and to find a (acceleration) we can use the formula a = d / 1/2 (t²) since Vi = 0 m/s.
a = d / 1/2 (t²)
= 15 m / 1/2 (3.25 s)²
= 15 m / 1/2 (10.56 s²)
= 15 m/ 5.28 s²
= 2.84 m/s²
Answer:
the acceleration of the jeepney is 2.84 m/s²
Explanation:
Hope it helps you, mark me as brainliest.
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We have the equation for distance traveled.
Distances (S) = u t + 1 a t²
–
2
u= initial velocity
t= time
a= acceleration
here:
u= o Since start from rest
a= 2.8 m/s²
t= 3.25 seconds
S= o x t 3.25 + 1/2 x 2.8 x 3.25²
=o +14.79 = 14.79 m
The jeep traveled 14.79 meters.
Explanation:
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