A bike moving on straight road cover 35 m in the 4th second and 40 m in 5 th second . What is it's initial Velocity ,,, and What is it Acceleration ? . ( Assume a = uniform ) .
Answers
Answered by
41
LHeya......!!!
Let S4 and S5 be the distance travelled ij 4th and 5th second respectively .
S4 = 35 m ,,,, S5 = 40 m .
From the equation of motion , the distance travelled in nth second is >>>
S(nth) = u + a(n-1/2)
35 = u + a ( 4-1/2 )............. i
40 = u + a( 5-1/2 )............ii
Now solving these 2 equations ,,,
35 = u + 7/2a
40= u + 9/2a
70= 2u + 7a
80= 2u + 9a
By further solving by simple mathematics .
【 a = 5 m/s^2 】
【u = 17.5 m/s ]
HOPE IT HELPS U ☺
Let S4 and S5 be the distance travelled ij 4th and 5th second respectively .
S4 = 35 m ,,,, S5 = 40 m .
From the equation of motion , the distance travelled in nth second is >>>
S(nth) = u + a(n-1/2)
35 = u + a ( 4-1/2 )............. i
40 = u + a( 5-1/2 )............ii
Now solving these 2 equations ,,,
35 = u + 7/2a
40= u + 9/2a
70= 2u + 7a
80= 2u + 9a
By further solving by simple mathematics .
【 a = 5 m/s^2 】
【u = 17.5 m/s ]
HOPE IT HELPS U ☺
Answered by
0
Answer:
17.5 m/s
Explanation:
==> 35 = u + a (4 - 1/2)
==> also 40 = u + a ( 4 - 1/2 )
==> 40 = u + 9/2 x a
==> 80 = 2u + 9a
a = 10/2 m/s2
u = 35/2 m/s
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