Question

a motorbike running at 5 m/s pick up a velocity of 30 m /s in 5s. calculate its acceleration and distance covered

Answer 1

heya mate
here is your answer

given,

initial velocity (v) = 5m/s
final velocity (u) = 30m/s
time = 5s

=>

acceleration = (v-u)/t

acceleration = (5-30)/5

acceleration = (-25)/5

acceleration = (-5)m/s

=> retardation = 5m/s


distance (s) = (v^2 - u^2) / 2a

distance = (5^2 - 30^2) / 2 × (-5)

distance = ( 25 - 900 ) / (-10)

distance = (-875) / (-10) = 87.5 m

Answer 2

Hello friend, 
Here's your answer. 

Given:
Initial velocity (u) : 5 m/s.
Final Velocity (v) : 30 m/s.
Time taken (t) : 5 seconds. 

Procedure:

Acceleration (a) = (v - u)/(t) 
=> a = (30 - 5)/(5) = 25/5 = 5 m/s².
Therefore acceleration = 5 m/s²

As we know the distance formula from Kinematics:

Distance (s) = ut + ½at²
So s = 5*5 + ½*5*5²
=> s = 25 + (125/2)
=> s = 25 + 62.5
Therefore distance = 87.5 metres. 

Hope my answer helps you.