Answer:
Explanation:
1) Name the two variables:
With that, the net rates of the boat down the river and upstrean are:
2) Now set the equations for the distance as a function of the times and the rates:
3) Set the system of equations:
4) Solve the system by elimination:
4 = 6.5 - c ⇒ c = 6.5 - 4 = 2.5
5) Results:
The rate of the motor boat in still water and the rate of the current is 6.5 mph and 2.5 mph respectively
Acceleration is rate of change of velocity.
[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]
[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]
a = acceleration ( m/s² )
v = final velocity ( m/s )
u = initial velocity ( m/s )
t = time taken ( s )
d = distance ( m )
Let us now tackle the problem!
Given:
distance traveled = d = 18 miles
time taken for travelling downstream = td = 2 hours
time taken for travelling upstream = tu = 4.5 hours
Unknown:
velocity of motor boat = vb = ?
velocity of current = vc = ?
Solution:
When motor boat traveled downstream , the velocity of motor boat was in the same direction to the velocity of current :
[tex]v_b + v_c = \frac{d}{t_d}[/tex]
[tex]v_b + v_c = \frac{18}{2}[/tex]
[tex]\boxed {v_b + v_c = 9}[/tex] → Equation 1
When motor boat traveled upstream , the velocity of motor boat was in the opposite direction to the velocity of current :
[tex]v_b - v_c = \frac{d}{t_u}[/tex]
[tex]v_b - v_c = \frac{18}{4.5}[/tex]
[tex]\boxed {v_b - v_c = 4}[/tex] → Equation 2
Next , Equation 1 and Equation 2 could be solved by using Elimination Method.
[tex](v_b + v_c) - (v_b - v_c) = 9 - 4[/tex]
[tex]2v_c = 5[/tex]
[tex]v_c = 5 \div 2[/tex]
[tex]v_c = \boxed {2.5 ~ \text{mph}}[/tex]
[tex]v_b + v_c = 9[/tex]
[tex]v_b + 2.5 = 9[/tex]
[tex]v_b = 9 - 2.5[/tex]
[tex]v_b = \boxed {6.5 ~ \text{mph}}[/tex]
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle
