Question

The initial velocity of an object is 15m/s^2 and its velocity is decreasing at 2m/s^2. What is the distance travelled by the object in 10s?

Answer 1

given already , 

u = 15 
t= 10 
s= ?

So, time passed before coming to rest.
v= u +at 
0= 15 -2(t)
 = -15= -2t
 = t= 7.5 sec
distance travelled before coming to rest --->
u = 15 m/s 
a = -2 m/s^2
t = 7.5 s
2as = v^2-u^2
2(-2)(s) = 0 - 225
s= 56.25

or 

s= ut + 1/2at^2
s= 15*7.5 + 1/2*-2*56.25 = 56.25

So,the object travels 56.25m displacement... before comig to rest

Now,The object changes direction because we are asked for 10 s but still now we found only for 7.5 s
t= 10 s given 
time left now = 10-7.5 = 2.5 s

s= ut + 1/2at^2 
= 0+ 1/2*-2*6.25
= -6.25m 
again so, distance travelled in other direction is 6.25 m

so, total distance = 56.25+6.25 = 62.5 m

whereas diplacement is 56.25-6.25 = 50 m

Answer 2

given,
⇒v=u+at
⇒0=15-2t 
⇒15/2=t
⇒7.5 = t
⇒u = 15 m/s
⇒a = -2 m/s^2
⇒t = 7.5 s
⇒2as = v^2 - u^2
⇒2(-2)(s) = -225
⇒-4s = -225
⇒s = -225/-2
⇒s = 56.25----------------------------(1)
now,
⇒t = 10s
left time = 10-7.5 = 2.5
therefore,
⇒s = ut+1/2at^2
⇒s = 0+1/2*-2*6.25
⇒s = -6.25
so, in direction in the opposite direction is 6.25

therefore distance traveled = 6.25+56.25 = 62.5

       i hope that this will help u...................^_^