Question

The relation between time and distance is distance(s)=3t+2t^2. Find Acceleration.

Answer 1

@Yahoo!

Answer 2

you are given distance
s=3t+t^2
rate of change of distance traveled is speed
and rate of change of velocity is acceleration

ds/dt = ds is the distance traveled in time dt

ds/dt= rate of change of s wrt time t....later on u'll come to know abt differentiation
here we are differentiating S wrt time t..

suppose y= x^n
then dy/dx= n x^(n-1)

u r being given...

s=3t+t^2
v= ds/dt= 3 t^(1-1) + 2 t^(2-1) = 3+2t
so v= ds/dt

Answer 3

again rate of change of velocity is acceleration....i.e. dv/dt=a
dv is the change in velocity in time dt

so u have v=3+2t
a= 0+ 2*t^(1-1)=2
so u got constant acceleration, which does not change with time.

Answer 4

\[\frac{ dS }{ dt }= V(velocity)=\frac{ d(3t+2t^2) }{ dt }=3+4t\] now
\[\frac{ dV }{ dt }= a(acceleration)=\frac{ d(3+4t) }{ dt }= 4\]

Answer 5

sorry i wrote
s=3 t +t^2
but u can solve for s= 3t +2 t^2

Answer 6

\[\Delta x=v_ot+\frac 12at^2\]\[\frac 12 a=2\]
\[a=4\]

Answer 7

The relation between time and distance is distance(s)=3t+2t^2. Find Acceleration.

v = d(3t+2t^2)/dx
v = 3+4t
a = d(3+4t)/dx
a = 0 + 4
a = 4