A motor bike covers 38.0 m in 12 s while slowing down to a final speed of 4.75 m/s. Find its original speed.
(Assume constant acceleration),
Answers
Answer:
(a) Here,
x
f
−x
i
=
2
1
(v
i
−v
f
)t becomes 40.0m=
2
1
(v
i
+280m/s)(8.50s)
Which yields v
i
=6.61m/s
(b) Similarly,
a=
t
v
f
−v
i
=
8.50s
2.80m/s−6.61m/s
=−0.448m/s
2
.
I HOPE IT IS HELPFUL TO YOU
Given : A motor bike covers 38.0 m in 12 s while slowing down to a final speed of 4.75 m/s.
(Assume constant acceleration),
To Find : original speed.
Solution:
S = 38 m
u = ?
v = 4.75 m/s
t = 12 s
a = ?
S = ut + (1/2)at²
=> 38 = 12u + (1/2)a(12)²
=> 38 = 12u + 72a
=> 19 = 6u + 36a
v = u + at
=> 4.75 = u + 12a
19 = 6u + 36a
4.75 = u + 12a => 14.25 = 3u + 36a
=> 4.75 = 3u
=> u = 4.75/3 m/s
Looks mistake in data
as u should be more than v as per question as slowing down mentioned
or direct formula
S = {( v + u)/2 } * t
=> 38 = ( 4.75 + u ) * 6
=> 6.33 = 4.75 + u
=> u = 1.5833 m/s
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