Question

Please answer these questions..

Answer 1

Answer:1st q: (a). 2nd q:(b)2,8,18,32

Explanation: The explanation for the questions, is as follows

Answer 2

(1)

Here the initial velocity, u = 0 m s^(-1), at t = 0 s.

After t seconds, the final velocity of the object will be,

v = u + at

v = at

Since the acceleration, a, is constant, average velocity is given by,

v_(av) = (u + v) / 2

v_(av) = a · t / 2

Alternate method:

At t = 0 seconds,

s(t) = ut + (at²) / 2

s(t) = u(0) + (a(0)²) / 2

s(0) = 0

After t seconds,

s(t) = ut + (at²) / 2

s(t) = at² / 2 [since u = 0].

Then, average velocity is,

v_(av) = [s(t) - s(0)] / (t - 0)

v_(av) = at² / 2t

v_(av) = a · t / 2

Hence (a) is the answer.

(2)

We have the second kinematic equation,

s = ut + (at²) / 2

Here, initial velocity, u = 0 m s^(-1), and acceleration, a = 4 m s^(-2). Then,

s = 4t² / 2

s = 2t²

If t = 1 s, s = 2(1)² = 2 m

If t = 2 s, s = 2(2)² = 8 m

If t = 3 s, s = 2(3)² = 18 m

If t = 4 s, s = 2(4)² = 32 m

Thus, the distance travelled by the car at the ends of 1, 2, 3 and 4 seconds are respectively 2, 8, 18 and 32 metres.

Hence (b) is the answer.

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