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# question02

last edited by 16 years, 10 months ago

# Question 2

## The use of a graphing calculator is allowed.

A particle moves along a line in such a way that at time t, 1 ≤ t ≤ 8, it's position is given by: (a) Write a formula for the velocity of the particle at time t.

(b) At what instant does the particle reach its maximum speed?

(c) When is the particle moving to the left?

(d) Find the total distance traveled by the particle from t = 1 to t = 8.

## Solution

### a) Write a formula for the velocity of the particle at time t. There are two ways on finding the velocity of this accumulation function. The long way moves anti-differentiating s(t) and then differentiating it once again. With that equation being a bit on the ugly side. We’ll go straight on the short, sweet, and easy way.

It is by using the Second Fundamental Theorum of Calculus

 If f is a continuous function on some interval (a, b) and an accumulation function A is defined by A(x) = (integral of) f(t) dt, then A’(x) = f(x)

Therefore, v(t) is equal to the underlying function.

V(t) = 1- x(cos x) - (ln x)(sin x)

### b) At what instant does the particle reach its maximum speed?

Always understand what you are being asked for. Speed is the absolute value of velocity. Therefore does not relate to the direction its going (+/-). Another thing it is asking for the maximum, meaning the endpoints are possibilities.

Graphing the velocity function and tracing for a maximum of the function, we get it has one on where x=3.7312. Tracing for a minimum of the function, we get that it has one on where x=1.2340 and x= 6.7002.

We evaluate the points and endpoints using the absolute value of the velocity function, V(t).

V(1) = 0.4597 <--Endpoint

V(1.2340) = 0.3938

V(3.7312)= 4.8334

V(6.7002) = 5.8964

V(8) = 0.1067 <-- Endpoint

Therefore, the particle reach its maximum speed at the instant when x=6.7002.

### c) When is the particle moving to the left?

We look at the velocity function we found in part a.

We find the roots of the function and whether its positively or negatively valued on either side of the root.

We remember our domain and only search for the roots in between.

v(t) 1 __+_____'___-_ 8

5.2056

Thus we know when our particle is moving to the left when it is negatively valued.

Therefore, the particle is moving to the left during (5.2036,8].

### d)Find the total distance traveled by the particle from t=1 to t=8.

Keyword: total distance

To find total distance we split the integral to the # of roots it has in the domain plus one. Our particular function has one root in the restricted domain. Therefore, we split the calculations into two.

S(5.2036) - (integral from 5.2036 to 8) v(t) dt

= 10.9923 - (-10.4675)

= 21.4599