Question

a truck has initial velocity of 20m/s and acceleration of -20m/s^2 .What distance will it covet before coming to a stop?​

Answer 1

Solution :

First let us find the time taken :

First Equation of Motion :

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

Where :

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• t = Time Taken

Using the First Equation of Motion and substituting the values in it, we get :

$:\implies \bf{v = u + at}$

$:\implies \bf{0 = 20 + (-20) \times t}$

[Note : Here the final Velocity is taken as negative because the truck is coming to rest]

$:\implies \bf{0 = 20 -20t}$

$:\implies \bf{- 20 = -20t}$

$:\implies \bf{\not{-} 20 = \not{-} 20t}$

$:\implies \bf{20 = 20t}$

$:\implies \bf{\dfrac{20}{20} = t}$

$:\implies \bf{1 = t}$

Hence, the time taken is 1 s.

Now , to find the distance traveled :

We know the second Equation of Motion i.e,

$\boxed{\bf{S = ut + \dfrac{1}{2}at^{2}}}$

Where :

• S = Distance
• a = Acceleration
• t = Time Taken
• u = Initial Velocity

Using the third Equation of Motion and substituting the values in it, we get :

$:\implies\bf{S = ut + \dfrac{1}{2}at^{2}}$

$:\implies\bf{S = 20 \times 1 + \dfrac{1}{2} \times (-20) \times 1^{2}}$

$:\implies\bf{S = 20 + \dfrac{1}{2} \times (-20) \times 1}$

$:\implies\bf{S = 20 + \dfrac{1}{2} \times (-20)}$

$:\implies\bf{S = 20 + (-10)}$

$:\implies\bf{S = 20 -10}$

$:\implies \bf{S = 10}$

$\boxed{\therefore \bf{S = 10\:m}}$

Hence, the distance covered is 10 m.

Alternative method :

We know the third Equation of Motion i.e,

$\boxed{\bf{v^{2} = u^{2} + 2aS}}$

Where :

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• S = Distance traveled

Now using the third Equation of Motion and substituting the values in it, we get :

$\boxed{\bf{0^{2} = 20^{2} + 2 \times (-20) \times S}}$

$:\implies \bf{0 = 20^{2} + 2 \times (-20) \times S}$

$:\implies \bf{0 = 20^{2} + 2 \times (-20) \times S}$

$:\implies \bf{- 20^{2} = 2 \times (-20) \times S}$

$:\implies \bf{- 400 = - 40S}$

$:\implies \bf{\not{-} 400 = \not{-} 40S}$

$:\implies \bf{\dfrac{400}{40} = S}$

$:\implies \bf{10 = S}$

$\boxed{\therefore \bf{S = 10\:m}}$

Hence, the distance covered is 10 m.