1.

A car accelerates from rest at a constant rate of 2m/s2 for sometime. Then it retards at a constant rate of 4m/s2 and comes torest it remains is motion for 6 sec then the total distance =?​pls tell the ans anyone

Answer»

Answer:

24 m

Explanation:

Time taken to travel first part of the journey (while accelerating) = t

Time taken to travel rest of the journey = 6 - t

Use equation of MOTION: v = U + at

In first part of journey

v = 0 + 2t

v = 2t .....(1)

In second part of journey

0 = v - 4(6 - t)

v = 4(6 - t) .....(2)

Divide equations (1) and (2)

v/v = 2t / (4(6 - t))

1 = t / (2(6 - t))

2(6 - t) = t

12 - 2t = t

3t = 12

t = 4 s

Then, from equation (1)

v = 2t

= 2 × 4

= 8 m/s

DISTANCE covered in first part

x = ut + 0.5at²

= (0 × 4) + (0.5 × 2 × 4²)

= 16 m

Distance covered in second part

y = vt + 0.5a't²

= (8 × 2) + (0.5 × -4 × 2²)

= 16 - 8

= 8 m

Total distance = 16 m + 8 m = 24 m


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