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.
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
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 :)
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.
To find the total distance traveled by the car, we need to consider the two stages of its motion: acceleration and deceleration.
1. Acceleration phase:
The car accelerates from 0 to 40 m/s in 5 seconds. We can use the formula for constant acceleration:
v = u + at
where:
v = final velocity (40 m/s)
u = initial velocity (0 m/s)
a = acceleration (unknown)
t = time (5 seconds)
Rearranging the formula, we have:
a = (v - u) / t
a = (40 - 0) / 5
a = 8 m/s²
Now, we can use the formula for distance traveled during constant acceleration:
s = ut + (1/2)at²
where:
s = distance (unknown)
u = initial velocity (0 m/s)
t = time (5 seconds)
a = acceleration (8 m/s²)
s = 0(5) + (1/2)(8)(5)²
s = 0 + 20
s = 20 meters
So, during the acceleration phase, the car travels 20 meters.
2. Deceleration phase:
The car decelerates from 40 m/s to 0 in 4 seconds. Since deceleration is the negative of acceleration, the deceleration value will be -8 m/s².
Using the same distance formula as before:
s = ut + (1/2)at²
where:
s = distance (unknown)
u = initial velocity (40 m/s)
t = time (4 seconds)
a = deceleration (-8 m/s²)
s = 40(4) + (1/2)(-8)(4)²
s = 160 - (1/2)(8)(16)
s = 160 - 64
s = 96 meters
During the deceleration phase, the car travels 96 meters.
Total distance traveled = distance during acceleration phase + distance during deceleration phase = 20 + 96 = 116 meters.
Therefore, the correct answer is not among the options provided.