a car accelerates from rest at 5 metre per second square and then retards to rest at 3 metre per second square. the maximum velocity of the car is 30 metre per second. what is the distance covered by the car over the entire journey????..
Answers
1st case: u = 0m/s
a = 5m/s^2
v = 30 m/s ( as the maximum velocity is achieved at the last stage
of acceleration.)
So, t = v-u/a
t = 30 - 0/ 5 seconds
t = 6 seconds
Thus, distance covered = 1/2 x 5 x 6^2 meters
d = 90 meters.
2nd case: u = 30m/s
v = 0m/s
a = -3 m/s^2
So, t = v-u/a
t = 0- 30/ -3 seconds
t = 10 seconds
Thus, distance covered = ut + 1/2 at^2
d' = 30 x 10 + 1/2 x (-3 x 10^2) meters
d' = 150 meters
Finally, Total distance covered = d + d'
= 90 + 150 meters
= 240 meters
Hence 240 meters will be covered by the car in its entire journey.