Question

A bicycle moves with a constant velocity of 5 km/h for 10 minutes and then decelerates at the rate 1 km/h2, till it stops. Find the total distance covered by the bicycle.

Answer 1

Find the distance covered at the first part and then the distance covered during the deceleration:
5km/hr = 1.389m/s
1km/hr2= 7.72 x 10^-5 m/s^2
10 minutes= 600 seconds

Kinematic equations:SUVAT
s-displacement, u-initial velocity, v-final velocity,a- acceleration, t-time
v=u + at
v2 = u2 + 2as
s=1/2(u+v)t
s=ut + 1/2at2

1)distance traveled before the start of deceleration
speed/velocity = 1.389m/s
if 1 second = 1.389 m
then 600s= 1.389 x 600 s
               = 833.33 meters traveled

2)distance traveled during deceleration:
v=0 , u=1.389m/s, a= 7.72 x10^-5 m/s2
v2=u2+2as
0=1.389^2 +2(-7.72x10^-5)s
0=1.93 - 1.543 x 10^-4 S
1.54x10^-4s= 1.93
s= 1.93/(1.54x10^-4)
 = 9.24 meters

Find total distance covered =  833.33m + 9.24 m
                                         = 842.57 m

Therefore the total distance covered by the bicycle is 842.57 meters.