Question

A particle moves along x axis with constant Acceleration and its x posting depend on time t as shown in the following graph (parabola ) then in interval 0 to 4 sec

• A. reletion between x- coordinate & time is $x=t-t^2/4$
• B. maximum x-coordinate is 1 m
• C. total distant traveled is 2 m
• D. average speed is 0.5 m/s

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A reletion between x- coordinate & time is $x=t-t^2/4$
B total distant traveled is 2 m
C maximum x-coordinate is 1 m
D average speed is 0.5 m/s

Let's take a general equation of parabola $x=a{t}^2+bt+c$ [where a,b,c are constants]

Now, at $t=0,x=0$ so, $c=0$

Also, $v=\frac{dx}{dt}=2at+b=$ slop of the graph

At t=0, slop is $tan{45}^{\circ }=1$

So at $t=0,v=1$

So, $2.a.0+b=1$

$or,b=1$

Now, at $t=4,x=0$

So, $0=a{.4}^2+1.4+0$ [putting value of $b,c$]

$or,a=-\frac{1}{4}$

After putting the values of $a,b.c$, we get

$x=t-\frac{t^2}{4}$

We will find the maximum x co-ordinate when $v$ will be $0$

now, $v=\frac{dx}{dt}=1-\frac{t}{2}=0$

$or,t=2$

So, $x_{max}=x_{t=2}=2-\frac{2^2}{4}=1m$

The total distance will be the sum of positive of values of $x$ from $t=0$ to $t=2$ and from $t=2$ to $t=4$ i.e

$|[x{]}_0^2+[x{]}_2^4|=1m+1m=2m$

Now, $Avergevelocity=\frac{Totaldistance}{Totaltime}=\frac{2}{4}m/s=0.5m/s$

So, $AllOptionsareCORRECT$