Question

A body is travelling with uniform acceleration and the final velocity is sqrt(180-7x) where x is the distance travelled by the body.Then the acceleration is?

Answer 1

I think you would do
\[ v= \frac{dx}{dt}= (180 -7x)^\frac{1}{2} \\
a= \frac{dv}{dt} = \frac{d^2x}{dt^2} = \frac{1}{2} (180-7x)^{-\frac{1}{2} }\cdot -7 \frac{dx}{dt}
\]
we can replace dx/dt to get
\[ a = -3.5 \cdot (180-7x)^{-\frac{1}{2} } \cdot (180-7x)^{\frac{1}{2} } \\ a= -3.5\]

Answer 2

Alternative Way :-
\[\large \bf v^2=u^2+2as\]
\[\large \bf v=\sqrt{u^2+2as}\]
and we have given
\[\large \bf v=\sqrt{180-7x}\]
compare them

Answer 3

we get,
\[\large \bf u^2=180\]
\[\large \bf 2ax=-7x\]
\[\large \bf a=\frac{-7}{2}=-3.5m/s^2\]

Answer 4

kya h be yeh?nahi samjha muhje

Answer 5

bhai,equation of 3rd motion likh

Answer 6

v^2 = u^2 + 2aS

Answer 7

tu usko aise bhi likh sakta h
\[\large \bf v=\sqrt{u^2+2as}\]

Answer 8

right?

Answer 9

ohk so on comparing
v^2 = u^2 + 2as with the given equation we get
v^2 = 180-7s

u^2 = 180

2as = -7s
2a = -7
s = -3.5ms^-2

Answer 10

yes