Question

The displacement (in metre) of a particle moving along x-axis is given by x = 18t + 5t

2

. Calculate :

(i) the instantaneous velocity at t = 2s,

(ii) average velocity between t = 2s and t = 3s.

(iii) instantaneous acceleration.​

Answer 1

Question:

The displacement (in metre) of a particle moving along x-axis is given by x = 18t + 5t². Calculate

(i) the instantaneous velocity at t = 2s,

(ii) average velocity between t = 2s and t = 3s.

(iii) instantaneous acceleration.

Solution

Given, x = 18t + 5t²

i) Instantaneous velocity at t = 2s,

v = dx/dt

v = d(18t + 5t²)/dt

v = 18 + 10t

Given, t = 2 sec. So,

v = 18 + 10(2)

v = 18 + 20

v = 38 m/s

ii) Average velocity between t = 2s and t = 3s.

Now,

x = 18t + 5t²

For t = 2 sec

x = 18(2) + 5(2)²

x = 36 + 5(4)

x = 36 + 20 = 56 m

For t' = 3 sec

x' = 18(3) + 5(3)²

x' = 54 + 5(9)

x' = 54 + 45 = 99 m

Distance = x' - x = 99 - 56 = 43 m

Time = t' - t = 3 - 2 = 1 sec

Average velocity = Distance/Time

= 43/1 = 43 m/s

iii) Instantaneous acceleration.

v = 18 + 10t (from above)

a = dv/dt

a = d(18 + 10t)/dt

a = 0 + 10 = 10 m/s

Answer 2

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The Question is Slightly Wrong.

Correct Question :

The displacement (in metre) of a particle moving along x-axis is given by x = 18t^2 + 5t.

Calculate :

(i) the instantaneous velocity at t = 2s,

(ii) average velocity between t = 2s and t = 3s.

(iii) instantaneous acceleration.

$\tt{ \purple{ \implies{v \: = \dfrac{dx}{t} = 36t + 18 }}}$

When T = 2,

Instantaneous velocity :

$\tt{ \purple{ \implies{v _{2} \: = \dfrac{dx}{t} = 36t + 18 = 90 \: m.s}}}$

$\tt{ \purple{ \implies{v _ {3} \: = \dfrac{dx}{t} = 36t + 18 = 126 \: m.s}}}$

Instantaneous Acceleration :

$\tt{ \orange{ \implies{a \: = \dfrac{dv}{t} = 36 \: m.s}}}$