Question

★ A car is moving with a speed of 10 m/s. When the brakes are applied, the car has a constant acceleration of -5 m/s² . What is the distance travelled before the car comes to rest? Also calculate the time taken by the car to come to rest.

✔ explanation should be in brief
✯no spam answers
✔answers will be deleted if,
✯irrelevant answer
✯copied
✯spammed
✯additional information required.​

_______________​​

Answer 1

Given :-

• Initial Velocity (u) = 10 m/s
• Final Velocity (v) = 0 m/s
• Time taken (a) = - 5 m/s²

To Find :-

• Distance travelled before the car comes to rest
• Calculate the time taken by the car to come to rest.

Solution :-

$\boxed{\sf a = v - \frac{u}{t}} \: ( \sf Formula \: of \: Acceleration )$

Where,

• a = Acceleration
• v = Final Velocity
• u = Initial Velocity
• t = Time taken

Substitute the values, we get

$\blue{\implies\sf - 5 = 0 - \frac{10}{t}} \\ \\ \green{\implies\sf - 5 = - \frac{10}{t}} \\ \\ \red{\implies\sf t = \frac{\cancel{- 10}}{\cancel{- 5}}} \\ \\ \orange{\implies\boxed{\sf t = 2}}</p><p>$

Hence, the time taken by the car to come to rest is 2 sec.

★ Using the third equation of motion ★

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

Where,

• v = Final Velocity
• u = Initial Velocity
• a = Acceleration
• s = Distance travelled

Substitute the values, we get

$\blue{\implies\sf 0^{2} - 10^{2} = 2 (- 5) \times s} \\ \\ \green{\implies\sf 0 - 100 = - 10s} \\ \\ \red{\implies\sf - 100 = - 10s} \\ \\ \orange{\implies\sf s = \frac{\cancel{ - 100}}{ \cancel{- 10}}} \\ \\ \gray{\implies\boxed{\sf s = 10}}$

Hence, distance travelled before the car comes to rest is 10 m.

Answer 2

★ QUESTION :

A car is moving with a speed of 10 m/s. When the brakes are applied, the car has a constant acceleration of -5 m/s² . What is the distance travelled before the car comes to rest? Also calculate the time taken by the car to come to rest.

GIVEN :

• Movement of car with speed [initial velocity of the car (u) ] = 10 m/s

• Constant acceleration of car [time taken (a)] = - 5m/s²

• [Final velocity (v)] of the car = 0 m/s

TO FIND :

• The time taken by car to come in rest = ?

• Distance travelled (before the car came in rest position) = ?

STEP -BY-STEP- EXPLAINATION :

First of all we will have to find the time taken by car to come in rest = ?

Let's find here :

$➠ a = - \: v \: \times \frac{u}{at}$

Substituting the values as per the given formula :

$➠ \: - 5 = 0 \times \frac{10}{t}$

$➠ \: - 5 = \frac{ - 10}{t}$

$➠ t = \frac{ - 10}{ - 5}$

Now, here the sign of - and - will be cut and it will change into the sign of +

$t = 2$

Therefore, The time taken by care to move in rest position = 2 second.

━━━━━━━━━━━━━━━━━━━━━━━━━━

Now, we will have to find distance travelled by car before it comes in rest position

The formula used for finding distance travelled by car before coming in rest, is as follows :

$➠ {v}^{2} \: - {u}^{2} \: = 2as$

Substituting the values as per the given formula :

$➠ \: {0}^{2} \: - {10}^{2} \:= 2( - 5) \: \times \: s$

$➠ \: 0 - 100 = - 10s$

$➠ \: - 100 = - 10s$

$➠ \: s = \frac{ - 100}{ - 10}$

Now, here the sign of - and - will change into +

$➠ \: s = 10$

Therefore, distance travelled by car before coming in the rest position = 10metre.

||━━━━━━━━━━━━━━━━━━━━━━||

⭐ ADDITIONAL INFORMATION ⭐

||━━━━━━━━━━━━━━━━━━━━━━||

★ What is the formula for solving these types of some, when the signs are use to be change ?

⟹ - and - = +

⟹ + and + = +

⟹ - and + = -

⟹ + and - = -

★ Important formula in physics (for solving the numericals of these types).

⟹ (u) = initial velocity (Speed)

⟹ (v) = Final velocity

⟹ (t) = Time taken (by object to cover the distance)

⟹ (a) = acceleration