Question

A body accelerates uniformly at 0.2m/s^2 and retards uniformly at 0.3m/s^2 . If the maximum velocity attained by it is 30m/s. Find the least time in which it can complete a journey of 6km.

Answer 1

This motion forms a trapezium whose area is the total distance covered.

Lets convert the total distance to meters.

6 × 1000 = 6000m

Acceleration = Change in velocity /t

0.2 = 30/t

0.2t = 30

t = 30/0.2 = 150 s

0.3 = 30/t

0.3t = 30

t = 30/0.3

t = 100 seconds

The car accelerates for 150 seconds and decelerates for 100 secons

These two motions form two triangles.

Area of the triangle for acceleration is :

0.5 × 150 × 30 = 2250 m

Area of triangle for deceleration :

0.5 × 30 × 100 = 1500m

The sum of this gives :

1500 + 2250 = 3750m

The remaining distance : 6000 - 3750 = 2250m

The car travels this distance at a constant speed.

The time taken is :

2250/30 = 75 seconds

The total time taken is :

75 + 100 + 150 = 325 seconds

Answer 2

Answer:

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