Question

A car is stopped at a traffic light. When the light turns green at t=0, a truck with a constant speed passes the car with a 20m/s velocity. The car uniformly accelerates to the velocity 35m/s in 20 seconds, the moves at a constant speed for 5 seconds. How long will it take for the car to catch up to the truck.

Answer 1

At
`t = (70 / 3) \; {\rm s}` (approximately
`23.3 \; {\rm s}`.)

Step-by-step explanation:

Note that the acceleration of the car between
`t = 0\; {\rm s}` and
`t = 20\; {\rm s}` (
`\Delta t = 20\; {\rm s}`) is constant. Initial velocity of the car was
`v_(0) = 0\; {\rm m\cdot s^(-1)}`, whereas
`v_(1) = 35\; {\rm m\cdot s^(-1)}` at
`t = 20\; {\rm s}\!`. Hence, at
`t = 20\; {\rm s}\!\!`, this car would have travelled a distance of:

`\begin{aligned}x &= ((v_(1) - v_(0))\, \Delta t)/(2) \\ &= \frac{(35\; {\rm m\cdot s^(-1)} - 0\; {\rm m\cdot s^(-1)}) * (20\; {\rm s})}{2} \\ &= 350\; {\rm m}\end{aligned}`.

At
`t = 20\; {\rm s}`, the truck would have travelled a distance of
`x = v\, t = 20\; {\rm m\cdot s^(-1)} * 20\; {\rm s} = 400\; {\rm m}`.

In other words, at
`t = 20\; {\rm s}`, the truck was
`400\; {\rm m} - 350\; {\rm m} = 50\; {\rm m}` ahead of the car. The velocity of the car is greater than that of the truck by
`35\; {\rm m\cdot s^(-1)} - 20\; {\rm m\cdot s^(-1)} = 15 \; {\rm m\cdot s^(-1)}`. It would take another
`(50\; {\rm m}) / (15\; {\rm m\cdot s^(-1)}) = (10/3)\; {\rm s}` before the car catches up with the truck.

Hence, the car would catch up with the truck at
`t = (20 + (10/3))\; {\rm s} = (70 / 3)\; {\rm s}`.