Question

a body moves with velocity 2 m/s for 5 s. then its velocity increases uniformly to 10 m/s in next 5 s. thereafter, its velocity begins to decrease until it comes to rest after 5 s
Find distance covered by body after 5 seconds and 10 seconds?

Answer 1

when body moves with velocity 2 m/s for 5 s
u =  2m/s 
t  =  5s 
v = 10 m / s  
a = v-u / t = 10 - 2 / 5  = 8 / 5 
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velocity begins to decrease until it comes to rest after 5 s
 u = 10 m / s
v = 0 m/s 
 a = -8/5 m/s^2 
t = 5 sec 
 s = ut + 1/2 at^2 
    = 50 - 20 = 30 m 
when t = 10 
s = 100 - 80 = 20 m 
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Answer 2

firstly body moves for 5 sec with velocity = 2 m/s
distance = speed × time taken
= 2 × 5 = 10 m

again, body moves with increasing velocity for next 5 sec
u = 2 m/s
V = 10 m/s
Use, formula,
a = ( v - u)/t
= (10 -2)/5 = 8/5 m/s²
a = 1.6 m/s²

V² = u² + 2aS
(10)² = (2)² + 2 × 1.6 × S
100 = 4 + 3.2× S
96 = 3.2 × S
S = 30 m


again, body retardated for next 5 sec
u = 10 m/s
V = 0 m/s
a = (0 -10)/5 = -2 m/s²
V² = u² +2aS
0² = 10² + 2×-2 × S
S = 25 m

hence,
distance after 5 sec = 10 m
and next 10 sec = (30 + 25) m = 55 m