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.
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Distance 1 = 5 * 10/60
5/6 km
First eq. of motion
V = u + at
0 = 5 - 1 * t
-5 = - t
T = 5 hours
2nd eq. of motion
Distance 2 = ut + 1/2at^2
= 5 * 5 + 1/2 (-1)(5)^2
= 25 - 25/2
= 25/2
Total distance = 5/6 + 25/2
= 80 / 6 km
Hope this helps you. Thanks
5/6 km
First eq. of motion
V = u + at
0 = 5 - 1 * t
-5 = - t
T = 5 hours
2nd eq. of motion
Distance 2 = ut + 1/2at^2
= 5 * 5 + 1/2 (-1)(5)^2
= 25 - 25/2
= 25/2
Total distance = 5/6 + 25/2
= 80 / 6 km
Hope this helps you. Thanks
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