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

Solution:

(given):

vi = 18.5 

vf = 46.1 

t = 2.47

d = ??

a = ??

Formula: a = (Delta v)/t

a = (46.1 m/s - 18.5 m/s)/(2.47 s)

a = 11.2 m/s2

d = vi*t + 0.5*a*t2

d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s2)*(2.47 s)2

d = 45.7 m + 34.1 m

d = 79.8 m

Answer 2

The both  accelerations in first and second case are given as follows -

Vi = 18.5 m/s and Vf = 46.1 m/s

and time for both accelerations is 2.47 m/s

      Now, a = v/t

                   = 46.1 - 18.5 / 2.47

                a = 11.2 m/s^2

               d = vi × t + 0.5 × a × t ^2

               d = 18.5 m/s × 2.47 + 0.5 × (11.2 m/s^2) × (2.47 s)^2

               d = 45.7 m + 34.1 m

               d = 79.8 m

   Therefore, the acceleration of car is 11.2 m/s^2 and distance (travelled) is 79.8 m.