Question

A vehicle accelerates from rest, travels at uniform velocity and then decelerates to rest all in a total time of 20 seconds. The time spent accelerating is equal to the time spent decelerating. For how long does it travel at uniform velocity?

Answer 1

The vehicle drives at uniform velocity for `14.14` s.

Explanation:

Total time = 20 s ⇒ `t_1 + t_2 + t_3 = 20` s
Total distance travelled, `s = vbar t = 5 × 20 = 100` m

Acceleration for `t_1` equals deceleration at `t_3` and magnitude of velocity change is the same. For `t_1` change in velocity is `|v-0| = v` and for `t_3` change in velocity is `|0-v| = v`. So `t_1 = t_3`.
`∴ t_1 + t_2 + t_3 = 2t_1 + t_2 = 20` s ①

Average velocity for `t_1` and `t_3` is `v/2` as acceleration and deceleration are constant and final / initial velocity is zero.

Area under acceleration and deceleration lines:
`s_1 = v/2 × t_1`
`s_3 = s_1`

Total area under the graph:
`s = s_1 + s_2 + s_3 = 2s_1 + s_2 = s(v/2 × t_1) + vt = v(t_1 + t_2)`
`⇒ 100 = v(t_1 + t_2)` ②

Substitute for `t_2` in equation ② from equation ①:
① `⇒ t_2 = 20 - 2t_1`
② `⇒ 100 = v(t_1 + 20 - 2t_1)`
`⇒ 100 = v(20 - t_1)` ③

Use equation of constant acceleration for `t_1`;
`v = u + at ⇒ v = 0 + 2 × t_1 ⇒ v = 2t_1` ④

Substitute for `v` in ③ from ④:
③ `⇒ 100 = 2t_1(20 - t_1)`
`⇒ -2t_1^2 + 40t_1 - 100 = 0`
`⇒ t_1^2 - 20t_1 + 50 = 0`

Solutions for the quadratic equation:
`t_1 = (-b ± sqrt (b^2-4ac))/(2a)`
Variables of the formula: a = 1, b = -20, c = 50
`t_1 = (20 ± sqrt (400-4 × 1 × 50))/(2 × 1)`
`t_1 = (20 ± 14.14/2`
`⇒ t_1 = 2.93` s or `17.07` s

Use equation ① to solve for `t_2`:
Equation ① is : `2t_1 + t_2 = 20` s
`⇒ t_2 = 20 - 2t_1`
So `t_2` is either `14.14` s or `-14.14` s. Obviously it cannot be `-14.14` s! So the solution is that the vehicle drives at uniform velocity for `14.14` s.