Question

A birds flies for 4s with a velocity of ( t-2) m/s in a straight line , where t is time in seconds. It covers a distance of

Answer 1

given

V= |t-2| 

dx/dt = |t-2|

dx = |t-2|dt

o∫xdx  = o∫4|t-2|dt

 x  = o∫2(2-t)dt  + 2∫4(t-2)dt

      =4

Answer 2

Answer:

The distance covered is 4 m.

Explanation:

Given that,

Time t = 4s

Velocity v = (t-2) m/s

We know that,

Distance :

The distance is product of the speed and time.

We know that,

The distance always positive so for two sec , the velocity depend on t will be given

$v = (2-t)\ m/s$

after 2 sec the velocity will be

$v=(t-2)\ m/s$

We know that,

Velocity :

The velocity is the rate of change of position of the particle.

$v = \dfrac{dx}{dt}$

$dx=v dt$

On integrating both side

$\int_{0}^{x}{dx}=\int_{0}^{4}{v}dt$

$\int_{0}^{x}{dx}=\int_{0}^{2}{(2-t)}dt+\int_{2}^{4}{(t-2)}dt$

$x=\int_{0}^{2}{(2-t)}dt+\int_{2}^{4}{(t-2)}dt$

$x=[2t]_{0}^{2}-[\dfrc{t^2}{2}]_{0}^{2}+[\dfrac{t^2}{2}]_{2}^{4}-[2t]_{2}^{4}$

$x=4-2+8-2-8+4=4\ m$

Hence, The distance covered is 4 m.