Calculating Distance Traveled by a Particle using Differentiation and Graphing

[takando12]

Homework Statement

The position of a particle along x-axis at time 't' is given by x=1+t-t².The distance traveled by the particle in the first second is.
A) 1m B) 2m C) 2.5m D) 3m

Homework Equations

The Attempt at a Solution

Differentiating x wrt t we get, v=1-2t and double differentiating x wrt to t we get a=-2m/s².
I plotted a V-t graph. The velocity will be 0 at t=0.5 sec , 1 at t=0sec and -1 at t=1 sec. Finding the area under the graph I got 0.5m. Where have I gone wrong? I got 2.5m when I substituted t=0.5 in x and multiplied it by 2. But why is the graph method not working?

 

[Nathanael]

Differentiating x wrt t we get, v=1-2t and double differentiating x wrt to t we get a=-2m/s².
I plotted a V-t graph. The velocity will be 0 at t=0.5 sec , 1 at t=0sec and -1 at t=1 sec. Finding the area under the graph I got 0.5m. Where have I gone wrong? I got 2.5m when I substituted t=0.5 in x and multiplied it by 2. But why is the graph method not working?

That's more complicated than it needs to be.

The function x(t) already gives you the position at each time, so there's no need to differentiate.

Finding the area under the graph I got 0.5m

How did you find the area under the curve? By integrating the velocity w.r.t. time? That just gets you back to the original function (except now you lost information about the initial position). Why differentiate just to integrate? It's redundant (and leaves more room for mistakes).

 

[takando12]

The function x(t) already gives you the position at each time, so there's no need to differentiate.

I differentiated it to find out when the velocity will be 0. Only after knowing that can i split the time period and use the x(t) to find the distance.

How did you find the area under the curve? By integrating the velocity w.r.t. time? That just gets you back to the original function (except now you lost information about the initial position). Why differentiate just to integrate? It's redundant (and leaves more room for mistakes).

I drew a v-t graph and there were two triangles , i added their area using 1/2*b*h*2.
[1/2*1/2*1] + [1/2*1/2*1]=0.5 m.

 

[haruspex]

I differentiated it to find out when the velocity will be 0. Only after knowing that can i split the time period and use the x(t) to find the distance.

Good so far.

I drew a v-t graph and there were two triangles , i added their area using 1/2*b*h*2.
[1/2*1/2*1] + [1/2*1/2*1]=0.5 m.

Good again.
It's the 2.5m answer that's wrong. The question asks for the distance travelled; the particle didn't start at 0.

 

[takando12]

Good so far.

Good again.
It's the 2.5m answer that's wrong. The question asks for the distance travelled; the particle didn't start at 0.

But 0.5 is not even in the options. And the answer key says it's 2.5m. Just to be clear, when x(t) =1 for t=0 sec. So when I calculate for 0.5 sec, the value of x I get is inclusive of the 1m in the beginning? So I should subtract x(0.5)-1= 1.25-1=0.25. And so the final answer 0.25*2=0.5 fits with the v-t graph.
So the options are wrong?

 

[Nathanael]

I differentiated it to find out when the velocity will be 0. Only after knowing that can i split the time period and use the x(t) to find the distance.

Sorry about that, I misinterpreted the question. (I thought it was asking for the displacement.)

But 0.5 is not even in the options. And the answer key says it's 2.5m. Just to be clear, when x(t) =1 for t=0 sec. So when I calculate for 0.5 sec, the value of x I get is inclusive of the 1m in the beginning? So I should subtract x(0.5)-1= 1.25-1=0.25. And so the final answer 0.25*2=0.5 fits with the v-t graph.
So the options are wrong?

Yes, the options are wrong, good job.

(As Haruspex pointed out, the answer is probably meant to be 2.5m, but that is a mistake. Your answer is correct.)

 

[takando12]

Sorry about that, I misinterpreted the question. (I thought it was asking for the displacement.)Yes, the options are wrong, good job.

(As Haruspex pointed out, the answer is probably meant to be 2.5m, but that is a mistake. Your answer is correct.)

Thank you so much sir or rather sirs.