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?

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Answer 1 · 2017-02-09T18:59:33

$15s$

Let the car cover three distances under accelerating, uniform and decelerating conditions $s_1,s_2 and s_3$ in time $t_1,t_2 and t_3$ respectively.

Using the kinematic equations
$v=u+at$ ......(1)
$v^2-u^2=2as$ ......(2)
$s=vt$ ......(3)

Accelerating
Let $v$ be velocity of car after time $t_1$ . From (1) we get

$v=0+5t_1$
$=>v=5t_1$ ......(4)

From (2) and (4)
$v^2-u^2=2alphas_1$
$(5t_1)^2-0^2=2xx5s_1$
$s_1=(5t_1)^2/10$
$=>s_1=2.5t_1^2$ .....(5)

Uniform motion
Using (3) and (4)

$s_2=(5t_1)t_2=5t_1t_2$ .....(6)

Decelerating
Using (1) and (4)

$0=5t_1-5t_3$
$=>t_1=t_3$ .....(7)

Using (2) and (4)

$0^2-(5t_1)^2=2(-5)s_3$
$=>s_3=s_1=2.5t_1^2$ .......(8)

We know that
Average Velocity $="Total displacement"/"Total Time"$

Inserting given values in SI units we get
$(s_1+s_2+s_3)/25=72xx1000/3600=20$

Using (5) (6) and (8) we get
$(2s_1+s_2)=500$
$(2xx2.5t_1^2+5t_1t_2)=500$
$=>(t_1^2+t_1t_2)=100$ ......(9)

Given is total time $=t_1+t_2+t_3=25$
Using (7)
$2t_1+t_2=25$
$=>t_2=25-2t_1$ .....(10)
Inserting this value is (9) we get

$(t_1^2+t_1(25-2t_1))=100$
$=>-t_1^2+25t_1=100$
$=>t_1^2-25t_1+100=0$

Roots of this quadratic are found by split the middle term and we get
$t_1=20 and 5$

Inserting these values one by one in (10) and ignoring other root which gives negative time $t_2$ we have the result

$t_2=25-2xx5=15s$

A resting car starts accelerating with α=5.0m/s^2α=5.0m/s^2 , then uniformly, and finally, decelerating at the same rate, αα, comes to a stop. The total time of motion equals t=25st=25s. The average velocity is 7272 kmph. For how long does the car move uniformly?