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(d) The particle moves in the positive direction when its velocity has a positive sign. You know the particle is at rest when and
, and because the velocity function is continuous, you need only check the sign of
for values on the intervals (0, 3) and (3, 6).
We have, for instance and
, which means the particle is moving the positive direction for
, or the interval (3, 6).
(e) The total distance traveled is obtained by integrating the absolute value of the velocity function over the given interval:
which follows from the definition of absolute value. In particular, if is negative, then
.
The total distance traveled is then 4 ft.
(g) Acceleration is the rate of change of velocity, so is the derivative of
:
Compute the acceleration at seconds:
(In case you need to know, for part (i), the particle is speeding up when the acceleration is positive. So this is done the same way as part (d).)
QUESTION 1
The given system of equations is
To solve by linear combination, we add equation (1) to equation (2) to get,
We divide through by 4 to obtain,
We put d=3 into equation (2) to get,
QUESTION 2
The given system is
4x + y = 5 ...eqn(1)
3x + y = 3 ...eqn(2)
To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.
This will give us,
This implies that,
Put x=3 into equation (1) to get,
The solution is
QUESTION 3
We want to solve the system;
a – 2b = –2 ....eqn(1)
2a + 2b = 14...eqn(2)
by linear combination.
We need to add equation (1) to equation (2) to eliminate b.
This implies that,
Simplify,
Divide both sides by 3 to get,
Put a=4 into equation (2) to obtain,
The ordered pair in the form (a, b) is
QUESTION 4
The given system of equations is
11x + 4y = 18 ...eqn(1)
3x + 4y = 2 ...eqn(2)
We subtract equation (2) from equation (1) to get,
Put x=2 into equation (2) to obtain,
This implies that,
The correct answer is (2,-1).
QUESTION 5
The given system is ;
2d + e = 8...eqn1
d – e = 4...eqn2
We add the two equations to eliminate e.
This implies that,
We divide both sides by 3 to get,
We put d=4 into equation (2) to get,
The solution is
The given system of equations is
To solve by linear combination, we add equation (1) to equation (2) to get,
We divide through by 4 to obtain,
We put d=3 into equation (2) to get,
QUESTION 2
The given system is
4x + y = 5 ...eqn(1)
3x + y = 3 ...eqn(2)
To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.
This will give us,
This implies that,
Put x=3 into equation (1) to get,
The solution is
QUESTION 3
We want to solve the system;
a – 2b = –2 ....eqn(1)
2a + 2b = 14...eqn(2)
by linear combination.
We need to add equation (1) to equation (2) to eliminate b.
This implies that,
Simplify,
Divide both sides by 3 to get,
Put a=4 into equation (2) to obtain,
The ordered pair in the form (a, b) is
QUESTION 4
The given system of equations is
11x + 4y = 18 ...eqn(1)
3x + 4y = 2 ...eqn(2)
We subtract equation (2) from equation (1) to get,
Put x=2 into equation (2) to obtain,
This implies that,
The correct answer is (2,-1).
QUESTION 5
The given system is ;
2d + e = 8...eqn1
d – e = 4...eqn2
We add the two equations to eliminate e.
This implies that,
We divide both sides by 3 to get,
We put d=4 into equation (2) to get,
The solution is