Linear Equation Word Problem

Question

David gets to work in 20 min when he drives his car. Riding his bike (by the same route) takes him 45 min. His average driving speed is 4.5 MOH greater than his average speed on his bike. How far does he travel to work?

Answer 1

Recall the formla

d = rt

Distance = rate times time

 

r_b = speed on bike

r_c=  r_b + 4.5 = speed in car

t_c = travel time in car = 20 min = 1/3 hour

t_b = travel time on bike = 45 min = 3/4 hour

d = d_c = d_b = distance travelled

 

dc = r_c t_c = (r_b + 4.5)(1/3)

d_b = r_b t_b = (r_b)(3/4)

Notice that we used the time expressed in hours since the rate (speed) is in miles per hour.

 

Since 

d = d_c = d_b; (r_b + 4.5)(1/3) = (r_b)(3/4)

(1/3)r_b + 1.5 = (3/4)r_b

1.5 = 3/2 = (5/12)r_b

3.6 mph = r_b

 

and since 

r_c= r_b + 4.5

r_c = 3.6 + 4.5 = 8.1 mph

 

Now that we have the rate, we can calculate the distance.

d = rt

So for the car, d = (8.1)(1/3) = 2.7 miles

and for the bike, d = (3.6)(3/4) = 2.7 miles

They should be equal.