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?

Question

1978 views around the world

You can reuse this answer
Creative Commons License

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#