In 5 seconds a car accelerate from 0 to 40 m/s. 20 sec after reaching 40m/s the car comes to a complete stop. if this deceleration takes 4 sec, how far has the traveled since its trip began?
a)100m/s
b)800m/s
-c)980m/s-
d)1140m/s
e)1700m/s
i got B but teach marked it wrong. can someone help me understand.
4 answers
i mean i got C
average 20 for first 5 so 100 m
go 40 for 20 s so 800 more
now brake average 20 for 4 = 80
980
so I agree
go 40 for 20 s so 800 more
now brake average 20 for 4 = 80
980
so I agree
The trick is to use average speed for a time of constant acceleration
v = Vi + a t
Vf = final v = Vi + a t
Vf -Vi = a t
so
a = (Vf -Vi)/t
d = Vi t + .5 a t^2
d = Vi t + .5 (Vf-Vi)t^2/t
d = Vi t -.5 Vi t + .5 Vf t
d = .5(Vi+Vf) t
BUT
.5(Vi+Vf) IS the average speed during acceleration :)
v = Vi + a t
Vf = final v = Vi + a t
Vf -Vi = a t
so
a = (Vf -Vi)/t
d = Vi t + .5 a t^2
d = Vi t + .5 (Vf-Vi)t^2/t
d = Vi t -.5 Vi t + .5 Vf t
d = .5(Vi+Vf) t
BUT
.5(Vi+Vf) IS the average speed during acceleration :)
does not do 40 for the whole 20 sec, but only for 20 - 4 = 16 s
so 100
+ 40*16 = 640
+ 80
= 820
..... not a choice though.
so 100
+ 40*16 = 640
+ 80
= 820
..... not a choice though.
1 answer