14. A racing car accelerates uniformly through three gears, changes with the following average speed:

Question

20 for 2.0 s
40 for 2.0 s
60 for 6.0 s

What is the overall average speed of the car?

Answer 1

average speed is distance / time

with initial velocity v, the distance s after time t is

s = vt + 1/2 at^2
and speed is v + at

at a=20, s = 0 + 1/2 * 20 * 2^2 = 40, v=20*2 = 40

at a=40, s = 40*2 + 1/2 * 40 * 2^2 = 160, v = 40+40*2 = 120

at a=60,
s = 120*6 + 1/2 * 60 * 6^2 = 1800

total distance: 40+120+1800 = 1960
total time: 2+2+6 = 10

avg speed: 1960/10 = 196

Answer 2

Sir but the and which u got is the same as i got but its not given in the options. here are the options

A. 12
B. 13.3
C. 40
D. 48
E. 37

Answer 3

Sir Steve I got the ans 46 but the correct ans is 48

S=vt+1/2at^2
in 1st gear
S=v*2+1/2*20*2^2
S=2v+40..........eq 1

in second gear

S=vt+1/2at^2
s=v*2+1/2*40*2^2
S=v+80.........eq 2

in third gear

S=vt+1/2at^2
S=v*6=1/2*60*6^2
S=6v+1080.......eq 3

substituting eq 3 in eq 2 we get
6v+1080=2v+80
6v-2v=80-1080
4v=-1000
v=-250

substituting the value of v in eq 1 we get
s=2v+40
s=2*-250+40
s=-460

vav=d/t

vav=460/10=46ms^-1

Answer 4

Using the definition of the average speed v (ave) = total distance/ time taken, we obtaine
d1 = v1•t1= 20•2 = 40 m
d2 = v2•t2 = 40•2 = 80 m
d3= v3•t3 = 60•6 =360 m
d= d1+d2+d3 = 40 + 80+360 =480 m
t= t1+t2+t3 = 2+2+6 = 10 s
v(ave) = d/t =480/140 =48 m/s

Answer 5

according to def v=d/t
d1 = v1•t1= 20•2 = 40 m
d2 = v2•t2 = 40•2 = 80 m
d3= v3•t3 = 60•6 =360 m
d= d1+d2+d3 = 40 + 80+360 =480 m
t= t1+t2+t3 = 2+2+6 = 10 s
v(ave) = d/t =480/140 =48 m/s