Question

A truck is stopped at a stoplight. When the light turns green, the truck accelerates at 3.0 m/s2. At the same instant, a car passes the truck going 12 m/s. Where and when does the truck catch up with the car?

Answer 1

The truck catches up to the car 96 meters from the stoplight, 8 seconds after the light turns green. This is found by equating the distance equations for the truck and car and solving for time.

Step-by-step explanation:

To find out where and when the truck catches up with the car, we first establish the equations of motion for both the truck and the car, starting from the moment the light turns green.

The car is already moving at 12 m/s and continues to move at this constant velocity, so its position over time is given by:

s (car) = vt

Where v = 12 m/s and t is in seconds. The truck, on the other hand, starts from rest and accelerates at 3.0 m/s², so its position over time is given by:

s (truck) = 1/2 at²

Where a = 3.0 m/s². To find when the truck catches up to the car, we set the position equations equal to each other:

1/2 at² = vt

Substitute the given values and solve for t:

1/2 × 3.0 m/s² × t² = 12 m/s × t

Cancel t from both sides (since at t=0, the truck hasn't caught up yet, it's the non-trivial solution we're interested in):

1/2 × 3.0 × t = 12

t = 8 seconds

To find out where they meet:

s = vt = 12 m/s × 8 s

s = 96 meters

Therefore, the truck catches up to the car 96 meters from the stoplight, 8 seconds after the light turns green.

Answer 2

The truck's position at time
`t` is

`x_(\rm truck)=\frac12\left(3.0(\rm m)/(\mathrm s^2)\right)t^2`

and the car's position is

`x_(\rm car)=\left(12(\rm m)/(\rm s)\right)t`

The two vehicle's meet when their positions are the same, at which point

`\frac12\left(3.0(\rm m)/(\mathrm s^2)\right)t^2=\left(12(\rm m)/(\rm s)\right)t`

`\implies\left(1.5(\rm m)/(\mathrm s^2)\right)t\left(t-8\,\mathrm s\right)=0`

`\implies\boxed{t=8.0\,\mathrm s}`