Question

A Tank having mass 10 tonnes is travelling with a velocity of 50 m/s. The

tankman applies emergency brakes and stops the tank in 10 seconds. Find the

deceleration produced in the tank.​

Answer 1

Answer:-

$\red{\bigstar}$ Deceleration produced by tank $\large\leadsto\boxed{\rm\purple{-5 m/s^2}}$

• Given:-

Mass of tank = 10 tonnes

→ 10 × 1000 $\dashrightarrow\bf\red{[\because 1 tonne = 1000 kg]}$

→ 10,000 kg

Initial velocity = 50 m/s

Final velocity = 0 m/s

Time = 10 sec.

• To Find:-

Deceleration of tank = ?

• Solution:-

We know,

$\large\boxed{\green{\bf v = u + at}}$

here,

v = Final velocity

u = Initial velocity

a = deceleration [negative acceleration]

t = time

Substituting in the values:-

➙ $\sf 0 = 50 + a \times 10$

➙ $\sf a \times 10 = -50$

➙ $\sf a = \dfrac{-50}{10}$

➙ $\bf a = -5 m/s^2$

Therefore, the declaration of the tank is -5 m/s² [negative as in opposite direction].

Answer 2

Given :-

☄ Initial velocity, u = 50 m/s

☄ Final velocity, v = 0 m/s

☄ Time taken, t = 10 sec.

To Find :-

Deceleration produced in the tank

Solution :-

by using equation of motion

➠ v = u + at

➠ 100 = 50 + (a)(10)

➠ 100 - 50 = 10a

➠ -50/10 = a

➠ a = -5m/s²

thus, the declaration of the tank is -5m/s²

|| Note :- If the speed decreasing with time then accerlation is negative. the negative accerlation is called declaration or retardation ||

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