# 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?

Aug 25, 2017

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 $\frac{v}{2}$ 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)

Substitute for ${t}_{2}$ in equation ② from equation ①:
⇒ t_2 = 20 - 2t_1
⇒ 100 = v(t_1 + 20 - 2t_1)
⇒ 100 = v(20 - t_1)

Use equation of constant acceleration for ${t}_{1}$;
v = u + at ⇒ v = 0 + 2 × t_1 ⇒ v = 2t_1

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

t_1 = (-b ± sqrt (b^2-4ac))/(2a)
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 : $2 {t}_{1} + {t}_{2} = 20$ 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.