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+3.09tt2.

At t=2.5, the displacement x(2.5)=1.96+3.092.52.52=3.435.

The velocity v is the derivative of displacement x with respect to time t, or v(t)=dxdt=3.092t. Thus, the velocity at time t=2.5 is v(2.5)=3.0922.5=1.91.

The acceleration a is the derivative of velocity v with respect to time t, or a(t)=dvdt=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+3.09tt2
so,
x=1.96+3.09tt2
Velocity is the rate of change of displacement, x
therefore, v=dxdt

v=d(1.96+3.09tt2)dt

v=(3.092t)ms

Now, acceleration is the rate of change of velocity, v
therefore, a=dvdt

a=d(3.092t)dt

a=2 ms2

Feb 10, 2018

Given, x=1.96+3.09tt2

at, t=2.50 ,s=1.96+(3.092.50)(2.50)2=3.435m

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

velocity =dxdt=3.092t

So,at t=2.50s its velocity is 3.09(22.50)=1.91ms

And,differentiating it one again we get, acceleration= dvdt=2ms2

ALTERNATIVELY

You can compare the equation with, s=ut12at2

So,you can rearrange as, x1.96=3.09t12×2×t2

So comparing we can say,it had an initial velocity of 3.09ms(at t=0) acceleration is 2ms2 and at t=0 it was 1.96m to the left of origin i.e along negative X axis.