Question

A car starts from rest at velocity 10m/s in 40secs. The driver applies a breakand slows down to 5m/s in 10secs. Find the acceleration in both the cases.​

Answer 1

Answer:

Acceleration in first case = 0.25 m/s²

Acceleration in second case = -0.5 m/s²

Explanation:

As per the provided information in the given question, we have :

In first case :

• Initial velocity (u) = 0 m/s

[ Whenever the body starts from rest its initial velocity is taken as 0.]

• Final velocity (v) = 10 m/s
• Time (t) = 40 s

In second case :

• Initial velocity (u) = 10 m/s
• Final velocity (v) = 5 m/s
• Time (t) = 10 s

$\underline{ \: \: \: \: \: \: \: \: \: \: \: \qquad\: \qquad \:\qquad \qquad \:\qquad\qquad \qquad \qquad\qquad}$

★ Calculating acceleration in first case :

By using the first equation of motion,

$\\ \longrightarrow \quad \pmb{\boxed{\sf {v = u + at}} }$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

Substituting values,

$\\ \longrightarrow \sf{\quad {10 = 0 + 40a }}$

$\\ \longrightarrow \sf{\quad {10 = 40a }}$

$\\ \longrightarrow \sf{\quad {\dfrac{10}{40} = a }}$

$\\ \longrightarrow \sf{\quad {0.25 = a }}$

$\\ \longrightarrow \bf{\quad \underline{ 0.25 \; m/s^2 = Acceleration_{(First \; case)} }}$

Therefore, acceleration in first case is 0.25 m/s².

$\underline{ \: \: \: \: \: \: \: \: \: \: \: \qquad\: \qquad \:\qquad \qquad \:\qquad\qquad \qquad \qquad\qquad}$

★ Calculating acceleration in second case :

By using the first equation of motion,

$\\ \longrightarrow \quad \pmb{\boxed{\sf {v = u + at}} }$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

Substituting values,

$\\ \longrightarrow \sf{\quad {5 = 10+ 10a }}$

$\\ \longrightarrow \sf{\quad {-5 = 10a }}$

$\\ \longrightarrow \sf{\quad {\dfrac{-5}{10} = a }}$

$\\ \longrightarrow \sf{\quad {-0.5 = a }}$

$\\ \longrightarrow \bf{\quad \underline{ -0.5 \; m/s^2 = Acceleration_{(Second \; case)} }}$

Therefore, acceleration in second case is -0.5 m/s².

More Information :

Equations of motion :

$\boxed{ \begin{array}{cc} \pmb{\sf{ \quad \: v = u + at \quad}} \\ \\ \pmb{\sf{ \quad \: s= ut + \cfrac{1}{2}a{t}^{2} \quad \: } } \\ \\ \pmb{\sf{ \quad \: {v}^{2} - {u}^{2} = 2as \quad \:}}\end{array}}$

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time
• s denotes distance

Answer 2

Answer:-

Firstly we calculate the acceleration in 1st case .

Initial velocity ,u = 0m/s

Final velocity ,v = 10m/s

Time taken ,t = 40 s

As we know that acceleration is defined as the rate of change in velocity .

a = v-u/t

Where,

a denote acceleration

v denote final velocity

u denote initial velocity

t denote time taken

Substitute the value we get

:\implies:⟹ a₁ = 10-0/40

:\implies:⟹ a₁ = 10/40

:\implies:⟹ a₁ = 1/4

:\implies:⟹ a₁ = 0.25 m/s²

Hence, the acceleration of the car is 0.25m/s²

_____________________________

Initial velocity ,u = 10m/s

Final velocity ,v = 5m/s

Time taken ,t = 10s

Now, calculating the acceleration in 2nd case .

• a₂ = v-u/t

Substitute the value we get

:\implies:⟹ a₂ = 5-10/10

:\implies:⟹ a₂ = -5/10

:\implies:⟹ a₂ = -0.5 m/s²

Here, negative sign show retardation

Hence, the acceleration in 2nd case is 0.5m/s².