Question

A race scooter is seen accelerating uniformly from 18.5 m/s to 46.1 m/s in 2.47 seconds. Determine the acceleration of the scooter and the distance travelled.

Answer 1

Answer

Given:-

Initial velocity of the  race scooter , u = 18.5 m/s

Final velocity of the  race scooter, v = 46.1 m/s

Time taken  by the  race scooter, t  = 2.47 sec

Acceleration of the race scooter = ???  

distance travelled of the race scooter = ???

1st case

In this we have ti use 1st equation of motion.

Acceleration a = v/t

                 a  = 46.1 - 18.5 / 2.47

                 a = 11.2 m/s²

 

2nd case

In this we have ti use 3rd equation of motion.

Distance,d = ut + (1/2)at²

 d= (18.5)×(2.47)+(1/2)×(11.2)×(2.47)²

 d= 45.7+34.17  

d= 79.87m

Hence, acceleration,a = 11.17m/s² and distance,d = 79.87m

Explaination:-

First equation of motion =  v = u + at

second equation of motion = s = ut + (1/2)²

Third equation of motion = v² = u² + 2as

Answer 2

$\huge\red{Answer}$

u=18.5 m/s

v=46.1 m/s

t=2.47 secs                                                                                                                 

a=(v-u)/t

= (46.1-18.5)/ 2.47

= 27.6/2.47= 11.17m/s^2

HOPE IT HELPS YOU !!