Question

A truck on a straight road starts from rest, accelerating at 2 m/s2 until it reaches a speed of 20 m/s. Then the truck travels for 20 s at constant speed until the brakes applied, are stopping the truck in a uniformmanner in an additional 5 s. (a) how long is the truck in motion? (b) What is the average speed of the truck for themotion described?

Answer 1

The entire motion took
`35\; {\rm s}`.

Average speed of the motion is
`(110/7)\; {\rm m\cdot s^(-1)}` (approximately
`16\; {\rm m\cdot s^(-1)}`.)

Step-by-step explanation:

The average speed of an object is equal to the ratio between total distance travelled and the duration of the motion. In this question, both of these quantities need to be found.

The average speed of this motion can be found in the following steps:

• Find the distance travelled when the truck was accelerating and decelerating.
• Find the total distance travelled during the motion, including the motion at constant speed.
• Divide the total distance travelled by the duration of the motion to find the average speed of the motion.

When the truck is accelerating
`a = 2\; {\rm m\cdot s^(-2)}`, divide the change in velocity by the acceleration (rate of change in velocity) to find the duration of the motion.

• The velocity of the truck was
  `u = 0\; {\rm m\cdot s^(-1)}` before the acceleration since the truck started from rest.
• The final velocity of the truck was
  `v = 20\; {\rm m\cdot s^(-1)}` after the acceleration.

Hence, the duration of the acceleration would be:

`\begin{aligned}t &= (v - u)/(a) = (20 - 0)/(2)\; {\rm s} = 10\; {\rm s}\end{aligned}`.

The combined duration of this motion would be:

`10\; {\rm s} + 20\; {\rm s} + 5\; {\rm s} = 35\; {\rm s}`.

While the truck was accelerating, since acceleration is constant, displacement
`x` of the truck would be:

`\begin{aligned}x &= (v^(2) - u^(2))/(2\, a) = (20^(2) - 0^(2))/(2\, (2))\; {\rm m} = 100\; {\rm m} \end{aligned}`.

Because the truck was moving in the same direction, the distance travelled would be equal to the magnitude of displacement:
`100\; {\rm m}`.

While the truck was decelerating:

• Duration of the motion is
  `t = 5\; {\rm s}`.
• Velocity before the acceleration was
  `u = 20\; {\rm m\cdot s^(-1)}`.
• Velocity after the acceleration was
  `v = 0\; {\rm m\cdot s^(-1)}` since the truck has stopped.

The displacement of the truck during this part of the motion would be:

`\begin{aligned} x &= \left((u + v)/(2)\right)\, t = \left((20 + 0)/(2)\right)\, (5)\; {\rm m} = 50\; {\rm m}\end{aligned}`.

Similarly, the distance travelled during this part of the motion is equal to the magnitude of displacement:
`50\; {\rm m}`.

While the truck travelled at a constant speed of
`20\; {\rm m\cdot s^(-1)}` for
`20\; {\rm s}` , the distance travelled was
`(20\; {\rm m\cdot s^(-1)})\, (20\; {\rm s}) = 400\; {\rm m}`. The distance travelled during the entire motion would be:

`(100\; {\rm m}) + (400\; {\rm m}) + (50\; {\rm m}) = 550\; {\rm m}`.

Divide the total distance travelled by the duration of the motion to find the average speed of the motion:

`\begin{aligned}(\text{average speed}) &= \frac{(\text{total distance})}{(\text{duration})} \\ &= \frac{550\; {\rm m}}{35\; {\rm s}} \\ &= (110)/(7)\; {\rm m\cdot s^(-1)} \\ &\approx 16\; {\rm m\cdot s^(-1)}\end{aligned}`.

In other words, the average speed of this motion would be
`(110/7)\; {\rm m\cdot s^(-1)}`, which is approximately
`16\; {\rm m\cdot s^(-1)}`.