Question

A 100kg vehicle moving with a speed of 20m/s is brought to rest in a distance 50 meters
find the acceleration and calculate the unbalanced force acting on a vehicle ​

Answer 1

Given:

Initial velocity of vehicle, u= 20 m/s

Final velocity of vehicle, v= 0 m/s

Distance covered by vehicle, s= 50m

To Find:

• Acceleration of given vehicle
• Unbalanced force acting on given vehicle

Solution:

Let the acceleration of given vehicle be 'a' m/s²

We know that,

• Force= mass×acceleration

• ​According to third equation of motion

$\pink{\boxed{\bf{v^{2}-u^{2}=2as}}}$

where,

v is final velocity of body

u is initial velocity of body

a is acceleration of body

s is displacement of body

So, on using above equation for given vehicle, we get

$\longrightarrow\mathrm{v^{2}-u^{2}=2as}$

$\longrightarrow\mathrm{0-(20)^{2}=2a(50)}$

$\longrightarrow\mathrm{-400=100a}$

$\longrightarrow\mathrm{100a=-400}$

$\longrightarrow\mathrm{a=\dfrac{-400}{100}}$

$\longrightarrow\mathrm{\green{a=-4\:m/s^{2}}}$

Here, negative sign tells us that acceleration is in opposite direction to motion of vehicle

Now, unbalanced force acting on given vehicle is

$\longrightarrow\mathrm{F=ma}$

$\longrightarrow\mathrm{F=100\times(-4)}$

$\longrightarrow\mathrm{\blue{F=-400\:N}}$

Here, negative sign tells us that force is acting in opposite direction of motion

Hence, acceleration is -4 m/s² and unbalanced force is -400 N.

Answer 2

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