Question

a truck and a car drive uniformly among the expressway from city a to city b. The truck leaves at 09:15 am and arrives at 1:15 pm. The car leaves at 10:00 am and arrives at 12:45 pm. At what times does the car overtake the truck? please help

Answer 1

Answer:

the car overtake the truck at time 11:40 am.

Step-by-step explanation:

We have both vehicules going at constant speed from city a to city b. The distance is unknown, but can be written as d.

We will express the time in hours (and decimals of hours).

The truck speed can be calculated estimating the time between arrival and start:

- The arrival time is 1.15 pm. This is t2=13.25.

- The starting time is 9:15 am. This is t1=9.25.

The truck took t2-t1=13.25-9.25=4 to go from city a to b.

The average speed is then:

$v_t=\dfrac{\Delta x}{\Delta t}=\dfrac{d}{4}$

We can write the equation for the position x(t) for the truck as:

$x(t)=x_0+v_t\cdot t=x_0+\dfrac{d}{4}t\\\\\\x(13.25)=x_0+\dfrac{d}{4}(13.25)=d\\\\x_0=d-3.3125d=-2.3125d\\\\\\x(t)=-2.3125d+0.25d\cdot t$

For the car we have:

- The arrival time is 12:45 am. This is t2=12.75.

- The starting time is 10 am. This is t1=10.

The car took t2-t1=12.75-10=2.75.

The average speed is then:

$v_c=\dfrac{\Delta x}{\Delta t}=\dfrac{d}{2.75}$

We can write the equation for the position x(t) for the car as:

$x(t)=x_0+v_c\cdot t=x_0+\dfrac{d}{2.75}t\\\\\\x(12.75)=x_0+\dfrac{d}{2.75}(12.75)=d\\\\x_0=d-4.6363d=-3.6363d\\\\\\x(t)=-3.6363d+0.3636d\cdot t$

The time at which the car overtake the car is the time when both vehicles have the same position:

$x(t)/d=-2.3125+0.25\cdot t = -3.6363+0.3636\cdot t\\\\-2.3125+3.6363=(0.3636-0.25)t\\\\1.3238=0.1136t\\\\t=1.3238/0.1136\approx11.65$

The car overtakes the truck at t=11.65 hours or 11:39 am.