Question

an object at rest at the origin begins to move in the +x direction with a uniform acceleration of 1 metre per second square for 4 seconds and then it continuous moving with a uniform velocity of 4metre per second in the same direction the x-t graph of objects motion will be​

Answer 1

Answer:

$t^{2}=2x$ ...... (1)

x=4t-8 ....... (2)

Step-by-step explanation:

The object starting from rest moves along +x direction. Assume that at t=0, x=0.

Now, initially it was moving with a constant acceleration of 1 m/sec² and continued for 4 sec, and then it moves with a uniform velocity of 4m /sec.

Hence, initially

$\frac{d^{2} x}{dt^{2} }=1$

⇒d($\frac{dx}{dt}$)=dt {Integrating both sides}

⇒ $\frac{dx}{dt}=t+c$ {Where c is the integration constant}

At t=0, $\frac{dx}{dt} = Velocity=0$

So, c=0.

Hence the equation becomes, $\frac{dx}{dt}=t$

⇒dx=t.dt {Again integrating both sides}

⇒x=$\frac{t^{2} }{2}+c'$ {Where c' is another integration constant}

Now , at t=0, x=0.

Hence, the above equation becomes, $x=\frac{t^{2} }{2}$

⇒ $t^{2}=2x$ ...... (1) is the initial equation. (Answer)

Now, after 4 secs, the velocity of the object will be,  

x=$\frac{1}{2} at^{2}$ (Since Initial velocity u =0}

⇒ x=$\frac{1}{2}(1)(4^{2})$= 8 m

Now in the second case, $\frac{dx}{dt} =4 m/sec$

dx=4dt {Integrating both sides}

x=4t+c" {Where c" is another integration constant}

Now, at t=4 sec , x=8 m,

Hence, c"=-8

So, the above equation becomes x=4t-8 ....... (2) (Answer)

See the diagram attached.