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:-

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

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• 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.

Answer 2

Given:-

• Car starts from rest at velocity 10m/s in 40 secs.

• The driver applies a breakand slows down to 5m/s in 10 secs.

To Find:-

• Acceleration in both the cases

Law used:-

• ${\bf{\boxed{First\:Law\:of\:Motion:\: v =u+at}}}$

Here,

• a = Acceleration

• v = Final Velocity

• u = Initial Velocity

• t = time

Solution:-

Let the acceleration before applying breaks be a and after applying be a'.

Case 1: When the breaks were not applied.

Using, v = u + at

Here,

• v = 10m/s

• u = 0m/s

• t = 40 sec

• a = a

Putting Values,

$\sf •\leadsto\:a =\dfrac{10-0}{40}$

$\sf •\leadsto\:a =\dfrac{10}{40}$

$\sf •\leadsto\:a =\dfrac{1}{4}$

${\bf{\boxed{•\leadsto\:a= 0.25ms^{-2}}}}$

Hence, The acceleration of the car before applying breaks is 0.25m/s²

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Case 2: When breaks were applied.

Again Using, v = u + at

Here,

• v = 5m/s

• u = 10m/s

• t = 10 sec

• a = a'

Putting Values,

$\sf •\leadsto\:a' =\dfrac{5-10}{10}$

$\sf •\leadsto\:a' =\dfrac{-5}{10}$

$\sf •\leadsto\:a' =\dfrac{-1}{2}$

${\bf{\boxed{•\leadsto\:a' = -0.5ms^{-2}}}}$

{negative sign show retardation}

Hence, The acceleration of the car after applying breaks is 0.5m/s².

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