Question

Position of a body changes according to equation x=t^2+2t-5. The average velocity of body in first three seconds is(x is in metre)
1) 4m/s
2) 5m/s
3) 20/3m/s
4) -20/3m/s

Answer 1

answer : option (2) 5m/s

explanation : position of particle changes according to equation , x = t² + 2t - 5

it is clear that, position is function of time. so, instantaneous velocity is the differentiation of x with respect to time,

e.g., dx/dt = d(t² + 2t - 5)/dt

or, v = 2t + 2

now, average velocity during (t1 ≤ t ≤ t2) is given by, <v> = $\frac{\int\limits^{t_2}_{t_1}{v}\,dt}{\int\limits^{t_2}_{t_1}\,dt}$

here, t1 = 0, t2 = 3s and v = (2t + 2)

so, <v> = $\frac{\int\limits^3_0{(2t+2)}\,dt}{\int\limits^3_0{dt}}$

= $\frac{[t^2+2t]^3_0}{3}$

= (3² + 6)/3

= 5 m/s

hence, average velocity = 5m/s

Answer 2

Answer:

5m/s

Explanation:

3^2+2×3÷3

9+6÷3

15÷3

=5m/s