A resting car starts accelerating with α=5.0m/s^2 , then uniformly, and finally, decelerating at the same rate, α, comes to a stop. The total time of motion equals t=25s. The average velocity is 72 kmph. For how long does the car move uniformly?

1 Answer
Feb 9, 2017

$15 s$

Explanation:

Let the car cover three distances under accelerating, uniform and decelerating conditions ${s}_{1} , {s}_{2} \mathmr{and} {s}_{3}$ in time ${t}_{1} , {t}_{2} \mathmr{and} {t}_{3}$ respectively.

Using the kinematic equations
$v = u + a t$ ......(1)
${v}^{2} - {u}^{2} = 2 a s$ ......(2)
$s = v t$ ......(3)

Accelerating
Let $v$ be velocity of car after time ${t}_{1}$. From (1) we get

$v = 0 + 5 {t}_{1}$
$\implies v = 5 {t}_{1}$ ......(4)

From (2) and (4)
${v}^{2} - {u}^{2} = 2 \alpha {s}_{1}$
${\left(5 {t}_{1}\right)}^{2} - {0}^{2} = 2 \times 5 {s}_{1}$
${s}_{1} = {\left(5 {t}_{1}\right)}^{2} / 10$
$\implies {s}_{1} = 2.5 {t}_{1}^{2}$ .....(5)

Uniform motion
Using (3) and (4)

${s}_{2} = \left(5 {t}_{1}\right) {t}_{2} = 5 {t}_{1} {t}_{2}$ .....(6)

Decelerating
Using (1) and (4)

$0 = 5 {t}_{1} - 5 {t}_{3}$
$\implies {t}_{1} = {t}_{3}$ .....(7)

Using (2) and (4)

${0}^{2} - {\left(5 {t}_{1}\right)}^{2} = 2 \left(- 5\right) {s}_{3}$
$\implies {s}_{3} = {s}_{1} = 2.5 {t}_{1}^{2}$ .......(8)

We know that
Average Velocity$= \text{Total displacement"/"Total Time}$

Inserting given values in SI units we get
$\frac{{s}_{1} + {s}_{2} + {s}_{3}}{25} = 72 \times \frac{1000}{3600} = 20$

Using (5) (6) and (8) we get
$\left(2 {s}_{1} + {s}_{2}\right) = 500$
$\left(2 \times 2.5 {t}_{1}^{2} + 5 {t}_{1} {t}_{2}\right) = 500$
$\implies \left({t}_{1}^{2} + {t}_{1} {t}_{2}\right) = 100$ ......(9)

Given is total time $= {t}_{1} + {t}_{2} + {t}_{3} = 25$
Using (7)
$2 {t}_{1} + {t}_{2} = 25$
$\implies {t}_{2} = 25 - 2 {t}_{1}$ .....(10)
Inserting this value is (9) we get

$\left({t}_{1}^{2} + {t}_{1} \left(25 - 2 {t}_{1}\right)\right) = 100$
$\implies - {t}_{1}^{2} + 25 {t}_{1} = 100$
$\implies {t}_{1}^{2} - 25 {t}_{1} + 100 = 0$

Roots of this quadratic are found by split the middle term and we get
${t}_{1} = 20 \mathmr{and} 5$

Inserting these values one by one in (10) and ignoring other root which gives negative time ${t}_{2}$ we have the result

${t}_{2} = 25 - 2 \times 5 = 15 s$