Question

A body is moving with constant velocity 12m/s whe is 32m behind a a cyclist the cyclist starts from rest and moves under constant acceleration 2m/s square After how much time the boy meets the cyclist ​

Answer 1

Answer :

Velocity of boy = 12m/s (constant)

Initial velocity of cyclist = zero

Acceleration of cyclist = 2m/s²

We have to find the time taken by cyclist to meet the boy$.$

Let both meet at time t.

In this interval of time, let distance covered by boy be d.

Cyclist have to cover distance of (d + 32) m in order to meet boy after time t.

★ Distance covered by boy :

➝ d = v × t

➝ d = 12t ____ (i)

★ Distance covered by cyclist :

➝ d + 32 = ut + 1/2 at²

➝ d + 32 = (0 × t) + 1/2 (2 × t²)

➝ d + 32 = t²

From the equation (i)

➝ 12t + 32 = t²

➝ t² - 12t - 32 = 0

➝ t² - (8 + 4)t - 32 = 0

➝ t² - 8t + 4t - 32 = 0

➝ t(t - 8) + 4(t- 8) = 0

➝ t = 8 or t = -4

Time can't be negative.

Therefore they meet after 8s.

Answer 2

$\fbox{given}$

velocity of body = 12m/sec

distance travelled= 32 m

initial velocity of cyclist= 0

acceleration= 2m/sec²

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$\fbox{to \: find}$

The time at which both the boy and the cyclist will meet .

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$\fbox{solution}$

$let \: boy \: and \: cyclist \: meet \: at \: time \: t.$

Let distance covered by the boy be d.

Distance that had to be covered by cyclist will be (32+d).

⛄Distance covered by the boy -

$distance= speed \times time$

d= 12 x t = 12t

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⛄ Distance covered by the cyclist-

[using s = ut + 1/2 at²]

d= 0 + 1/2 x 2 x t²

d+ 32 = t²

putting value of d in d+ 32 = t²

12t + 32= t²

t²-12t-32= 0

By solving the quadratic equation, we get :

(t+4) (t-8)= 0

t= 8 , -4

since time cannot be negative so the boy and the cyclist will meet at 8 second.

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