A scooter acquires a velocity of ` 36 km//h` in `10 seconds` just after the start . It takes `20` seconds to stop . Calculate the acceleration in the two cases.

Explanation

Case I. Here , initial velocity , `u = 0`
final velocity , `v = ( 36 km)/( h) = ( 36 xx 1000 m)/( 60 xx 60 s) = 10 m//s`
time taken , `t = 10 s , a = ?`
As `a = ( v - u)/( t) , a = ( 10 - 0)/( 10) = 1 m//s^(2)`
Case II. Here , ` u = 36 km//h = 10 m//s`
As the scooter stops , `v = 0 , t = 20 s: a= ? `
Again ,as `a = ( v - u)/( t) , a( 0 - 10) /( 20) = - 0.5 m//s^(2)`
Negative value of `a` shows that the motion is retarded in case II.