Question

A car travels at a speed of 54 km per hour application of brakes it stopped in 10 seconds what is the acceleration of car and and what distance will be travelled before stopping​

Answer 1

Given :

• Initial velocity (u) = 54 km/h
• Final velocity (v) = 0 km/h
• Time interval (t) = 10 seconds

To Find :

• Acceleration of the car
• Distance travelled before stopping

Formulae Used :

• v = u + at
• v² - u² = 2as

Explanation :

A car is moving with a velocity of 54 km/h, then brakes are applied to the car and car stops after 10 seconds, breaks are applied it means that the car comes to rest means initial velocity will be 0 km/h.

Convert the velocity to m/s from km/h by multiplying it by 5/18.

⇒Initial velocity (u) = 54 * 5/18 = 15 m/s

• Initial velocity (u) = 15 m/s
• Final velocity (v) = 0 m/s
• Time (t) = 10 s

We have to use Kinematics equation in this question, first by using 1st equation we have to find value of acceleration. And after that using 3rd equation we will find out distance travelled.

$\bigstar \: \: \boxed{\sf{v \: = \: u \: + \: at}} \\ \\ \bigstar \: \: \boxed{\sf{v^2 \: - \: u^2 \: = \: 2as}}$

Where,

• v is final velocity
• u is initial velocity
• a is acceleration
• t is time interval
• s is distance travelled

Solution :

$\implies \sf{v \: = \: u \: + \: at} \\ \\ \implies \sf{15 \: = \: 0 \: + \: 10a} \\ \\ \implies \sf{a \: = \: \dfrac{15}{10}} \\ \\ \implies \sf{a \: = \: 1.5}$

$\therefore$ Acceleration of the car is 1.5 m/s²

________________________

Now, distance travelled is :

$\implies \sf{v^2 \: - \: u^2 \: = \: 2as} \\ \\ \implies \sf{15^2 \: - \: 0^2 \: = \: 2 \: \times \: 1.5 \: \times \: s} \\ \\ \implies \sf{225 \: = \: 3s} \\ \\ \implies \sf{s \: = \: \dfrac{225}{3}} \\ \\ \implies \sf{s \: = \: 75}$

$\therefore$ Distance travelled is 75 m

Answer 2

Given that ,

• Final velocity (u) = 0 m/s
• Initial velocity (v) = 54 km/hr or 15 m/s
• Time (t) = 10 sec

We know that , the first equation of motion is given by

$\star \: \: \sf v = u + at$

Thus ,

$</p><p>\Rightarrow \sf 15 = 0 + a × 10 \\ \\</p><p>\Rightarrow \sf a = \frac{15}{10} \\ \\ </p><p>\Rightarrow \sf </p><p>a = 1.5 \: \: m/ {s}^{2} </p><p>$

$\therefore \sf \bold{ \underline{The \: acceleration \: is \: 1.5 m/ {s}^{2} }}$

And the third equation of motion is given by

$\star \: \: \sf {(v)}^{2} - {(u)}^{2} = 2as$

Thus ,

$</p><p>\Rightarrow \sf {(15)}^{2} + {(0)}^{2} = 2 \times 1.5 \times s \\ \\</p><p>\Rightarrow \sf 225 = 3 \times s \\ \\</p><p>\Rightarrow \sf s = \frac{225}{3} \\ \\</p><p>\Rightarrow \sf s = 75 \: \: m$

$\therefore \sf \bold{ \underline{The \: distance \: is \: 75 \: m}}$