Question

A particle has a displacement of 40m along OX in the first 5s and 30m perpendicular to OX in

the next 5s. The magnitude of average velocity of the particle during the time internal of 10s is

a) 6 ms–1 b) 7 ms–1 c) 8 ms–1 d) 5 ms–1 [ ​

Answer 1

Answer:

A body covers 20m in 2 sec and 80 m in 4 seconds from the start along a straight line. If it is moving with constant acceleration, how much distance will it cover

Let the initial velocity of the body be u m/s; and the constsnt acceleration be a m/s². Then displacemet (distance covered) of the body after t second is given by

s = u t + ½ a t² ———————————-(1)

The body covers 20 m in 2s and 80 m in 4 s.

20 = u×2 + ½ a 2²; => 20 = 2 u + 2 a

u + a = 10 —————————————(2)

80 = u×4 + ½ a 4²; => 80 = 4 u + 8 a; this gives,

u + 2 a = 20 ——————————————(3)

Solving (2) and (3) for u and a we get

u = 0 m/s and a = 10 m/s² ————————(4)

To obtain the distance covered in the next 4s we need to find the velocity at the beginning of this interval using,

v = u + a t => v = 0 m/s +

Answer 2

Explanation:

v1= 40/5=8

v2=30/5=6

v1+v2/2

8+6/2

14/2

7ms

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