Application of differentiation

Question

The velocity v m s –1 of a particle at time t seconds, is given by v = 2t + cos2t, for 0 ≤ t ≤ 2.

(a) Write down the velocity of the particle when t = 0.

 

When t = k, the acceleration is zero.

(b) (i) Show that k = 4 π .

 

(ii) Find the exact velocity when t = 4 π .

 

(c) When t < 4 π , t v d d > 0 and when t > 4 π , t v d d > 0. Sketch a graph of v against t.

 

(d) Let d be the distance travelled by the particle for 0 ≤ t ≤ 1.

 

(i) Write down an expression for d.

(ii) Represent d on your sketch.

Answer 1

I am not sure how to do sketches on here but I can help with the other parts.

 

(a) Since we are given the velocity as a function of time, to find the velocity at t = 0, just subsitiute 0 for t.

v = 2t + cos2t

  = 2(0) + cos2(0)

  = 0 + 1 = 1

 

(b) an acceleration function can be found by taking the derivative of the velocity function

dv/dt = 2 - 2sin2t

to find when the acceleration is 0 set dv/dt = 0

0  = 2 - 2sint2t

-2 = -2sin2t

2  = 2sin2t

2  = 2sin(2*4π)

2  = 2sin8π

2  = 2*(1)

2  = 2 therefore when t = k = 4π the acceleration is 0

 

to find the velocity at 4π just substitute that back into the original velocity function

 

v = 2t + cos2t

   = 2(4π) + cos(2*4π)

   = 8π + 1

 

(c) not sure how to add a sketch to here but you can use a graphing calculator or pick some points to help you sketch

the graph.

 

(d) Talking about the distance traveled is asking about the displacement or position function.  That can be obtained by 

integrating your velocity function and evaluating it from 0 <= t <= 1.

I don't know how to make the integral sign on here but the antiderivative would be

t² + (sin2t)/2 evaluated from 0 to 1;

1² + (sin(2*1))/2 - (0² + (sin(2*0))/2

1 + (sin(2))/2 - 0

1 + 0.5*sin2

(i) the expression would be t² + (sin2t)/2 + C (As the most generic) 

(ii) Again I don't know how to sketch on here sorry.

 

One problem I have unless I am missing something is when being asked to evaluate the acceleration and velocity at t = 4π.  That would be out of the initial interval for t given.