Starting from rest, an object is uniformly accelerated to a speed of 60m/s in 20s. The object travels with this speed for next 40s and is then brought to rest by a uniform retardation in the next 30s.Sketch velocity time graph and calculate the acceleration, retardation and the total distance travelled.
Hint: acceleration =rate of change of velocity.  distance covered =area under velocity-time graph .

Acceleration during initial phase of travel is determined from the following equation of motion

 

v = u + ( a t  )

 

where u = initial speed = 0  and  v = 60 m/s is speed attained with uniform acceleration a in t = 20 s

 

a = v / t = 60 / 20 = 3 m/s²

 

Distance S₁ travelled in initial phase , S₁ =  (1/2) a t² = 0.5 × 3 × 20 × 20 = 600 m

 

Distance S₂ travelled in uniform speed 60 m/s for 40 s is calculated as

 

S₂ = 60 × 40 = 2400 m

 

Retardation during last phase of travel is determined from the following equation of motion

 

v = u - ( a_r t  )

 

where u = initial speed = 60 m/s   and  v = 0  m/s is final speed after t = 30 s

 

a_r = u / t = 60 / 30 = 2 m/s²

 

Distance S₃ travelled while retardation is calculated as follows

 

S₃ = ( u t ) - [ (1/2) a t² ]  =  ( 60 × 30 ) - ( 0.5 × 2 × 30 × 30 ) = 900 m

 

Total distance S = S₁ + S₂ + S₃ = 600 + 2400 + 900 = 3900 m