dog initially running at a speed of 2 metre per second accelerates at a constant rate of 1.5 m per second square for five seconds calculate 1. the distance covered by the dog from its initial position. 2 . rhe final velocity of the dog.
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Answered by
3
S= Vf^2 - Vi^2 / 2 a
Vf = vi + at
Vf = 2 + 1.5 × 5 = 9.5 m/s
S= (9.5)2 - (2)^2 / 2× 1.5 = 28.75 m
Vf = vi + at
Vf = 2 + 1.5 × 5 = 9.5 m/s
S= (9.5)2 - (2)^2 / 2× 1.5 = 28.75 m
jlo1527:
How? I mean the working...
Apply S= (Vf)^2 - (Vi)^2 / 2 a
Vf= vi + at
Vf = 2 + 1.5 × 5 = 9.5
S = (9.5)^2 - (2)^2 / 2×1.5 = 28.75 m
Thanx.. but then, is mine wrong??
Your calculations are wrong :')
10 + 18.75 = 28.75 not 19.75
yup i corrected it... thanx
so both the formulas can be applied..
Answered by
8
Hi
your answer can be simplified by some formulas :
1.
S=ut + 1/2 at^2
S= 2*5 + 1/2 *1.5*5*5
S = 10 + 18.75
S = 28.75 (distance)
2.
V= u+at
V= 2+ 1.5*5
V= 9.5 ( final velocity)
your answer can be simplified by some formulas :
1.
S=ut + 1/2 at^2
S= 2*5 + 1/2 *1.5*5*5
S = 10 + 18.75
S = 28.75 (distance)
2.
V= u+at
V= 2+ 1.5*5
V= 9.5 ( final velocity)
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