Question

A car is moving with a velocity of 20m/s. When the driver applies brake, the car retards with 2m/s. In what time will the car come to rest?​

Answer 1

Answer:

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Answer 2

In 10 seconds the car will come to rest.

Explanation :

Here, we are provided in the question that a car is moving with a velocity of 20m/s that means it's initial velocity is 20 m/s. Then, the driver applies brake, so it's final velocity is 0 m/s. After applying brakes the car retards with 2m/s, retardation is 2m/s².

Considering the statements stated in the question, we have :

• Intial velocity (u) = 20 m/s
• Final velocity (v) = 0 m/s [It applies brake]
• Retardation = 2 m/s²

We have to find in what time will the car come to rest. Here, we'll be using the first equation of motion. By using the first equation of motion,

$\bigstar \: \boxed{\sf {v = u + at}}$

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

Here, acceleration (a) is - 2m/s².

Retardation is nothing but acceleration with the negative sign. If acceleration will be positive then retardation will be negative and if acceleration will be negative then retardation will be positive. So, here if retardation is +2 m/s² then acceleration will be -2 m/s².

Now, substituting the values in the first equation of motion.

$\longrightarrow \sf {v = u + at }$

$\longrightarrow \sf {0 = 20 + ( -2t ) }$

$\longrightarrow \sf {0 = 20 - 2t }$

$\longrightarrow \sf {0 + 2t = 20}$

$\longrightarrow \sf { 2t = 20}$

$\longrightarrow \sf {t = \dfrac{20}{2} }$

$\longrightarrow \boxed{ \sf {t = 10 \: seconds}}$

Therefore, in 10 seconds the car will come to rest.

More related formulae :

Three equations of motion ::

• v = u + at
• s = ut + $\sf { \dfrac{1}{2}at^2}$
• v² - u² = 2as

Here,

• v denotes final velocity.

• u denotes initial velocity.

• a denotes acceleration.

• s denotes distance.

• t denotes time.