Would someone please help me with this physics problem?

Question

A particle moving at 10 m/s reverses its direction to move at 20 m/s in the opposite direction. If its acceleration is -10 m/s^2, what is the total distance that it travels?

A 15 m
B 20 m
C 25 m
D 30 m

Please don't just give me an answer. I need an explanation. Give me the reasoning. Thanks.

Answer 1

u=10
a= -10
for time taken by it to come to rest :
v=0 hence u+at=0
t=-u/a=1sec
distance covered in this one second s=ut+1/2at^2
=10(1)+1/2(-10)(1)^2
=10-5=5m
now it is accelarating in opposite direction hence time taken
for its velocity to become 20m/s
v=20 ,u=0(it starts from rest)
v=u+at
20=0+10t
t=2sec
distance covereds=0+1/2at^2=1/2*10*2^2=20m
total distance traveled=20+5=25m

Answer 2

Before you start use KISS principal and watch the signs. DO NOT JUST USE FORMULAS.

The Ans is definitely (A) and this is why

The equation for velocity is
V(t)=-V0 + at where
V(t) - velocity as a function of time
a - acceleration
t - time
V0 - initial velocity

For our motion due to the reversal we have to write
V(t)=- -V0 + at
then
t=(V(t)+V0)/a
t=(20+10)/10 =3 sec
now

S=-Vot+ .5at^2 where
S distance
a - acceleration
t -time
V0 - initial velocity
S=-10 (3 )+0.5(10)(3)^2= -30 + 45m
S=15m


The reason others made a mistake they did not derive the equations. had they they would have
S=(V(t)^2-V0^2)/(2 a) [it is minis initial velocity because of reversal!]
S=(20^2 - 10^2)/(2 x 10)=15m (that minus sign did them in)
Let me know if you have any questions

Answer 3

v^2=u^2 +2ad
a=-10;
u = 10m/s
v = 0m/s (it changes direction after that)

0 = 100 + 2(-10)(d)
d= 5m
second section
v=-20
u=0
a=-10
400 = 0 + 2(-10)(d)
d= -20m

total distance traveled = |5| + |-20| = 25m
C