What is its velocity? What is its acceleration?

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3 Answers
Feb 10, 2018

x(2.5)=3.435
v(2.5)=-1.91
a(t)=-2

Explanation:

It is given that the displacement x at time t is given by x(t)=1.96+3.09t-t^2.

At t=2.5, the displacement x(2.5)=1.96+3.09*2.5-2.5^2=3.435.

The velocity v is the derivative of displacement x with respect to time t, or v(t)=dx/dt=3.09-2t. Thus, the velocity at time t=2.5 is v(2.5)=3.09-2*2.5=-1.91.

The acceleration a is the derivative of velocity v with respect to time t, or a(t)=(dv)/dt=-2. Since it stays constant regardless of the time, the acceleration at time t=2.5 is -2.

Feb 10, 2018

The particle moves according to the equation, 1.96 + 3.09t - t^2
so,
x=1.96 + 3.09t - t^2
Velocity is the rate of change of displacement, x
therefore, v=dx/dt

v=(d(1.96 + 3.09t - t^2))/dt

v=(3.09-2t)m/s

Now, acceleration is the rate of change of velocity, v
therefore, a=dv/dt

a=(d(3.09-2t))/dt

a=-2 m/s^2

Feb 10, 2018

Given, x=1.96+3.09t-t^2

at, t=2.50 ,s=1.96+(3.09*2.50)-(2.50)^2=3.435 m

differentiating the equation w.r.t time,we get,

velocity =(dx)/(dt) = 3.09 -2t

So,at t=2.50s its velocity is 3.09 -(2*2.50)=-1.91 m/s

And,differentiating it one again we get, acceleration= (dv)/(dt) = -2 m/s^2

ALTERNATIVELY

You can compare the equation with, s=ut -1/2 at^2

So,you can rearrange as, x-1.96=3.09t - 1/2 ×2×t^2

So comparing we can say,it had an initial velocity of 3.09 m/s(at t=0) acceleration is -2 m/s^2 and at t=0 it was 1.96 m to the left of origin i.e along negative X axis.