Question

Only for Physics specialists..

A car is moving at 20 m/s. The driver applies the breakers which causes a retardation of 4 m/s^2. Find thr distance travelled by the car before it stops, if reaction time of driver is 0.4 seconds.

Happy times..​

Answer 1

Answer- The above question is from the chapter 'Kinematics'.

Some important terms and formulae:-

1. Velocity- It is the displacement per unit time.

S.I. Unit of Velocity- m/s

It is a vector quantity as it possesses magnitude and direction.

2. Acceleration- It is the rate of change of velocity.

S.I. Unit of Acceleration- m/s²

It is also a vector quantity.

Negative acceleration is called retardation.

3. Distance- It is the path length transversed by an object.

S.I. Unit of Distance- m

It is a scalar quantity.

4. Displacement- It is the shortest distance between the initial and final point.

S.I. Unit of Displacement- m

It is a vector quantity.

5. Equations for uniformly accelerated motion-

Let u = Initial velocity of a particle

v = Final velocity of a particle

t = Time taken

s = Distance travelled in the given time

a = Acceleration

1) v = u + at

2) s = $\frac{1}{2}$ at² + ut

3) v² - u² = 2as

Given question: A car is moving at 20 m/s. The driver applies the breakers which causes a retardation of 4 m/s². Find the distance travelled by the car before it stops, if reaction time of driver is 0.4 seconds.

Answer: From 1st point to 2nd point,

Initial Velocity (u) = 20 m/s

Time (t) = 0.4 s

Distance (s₁) = ut (since there is no acceleration involved.)

s₁ = 20 × 0.4 = 8 m

From 2nd point to 3rd point,

u = 20 m/s

v = 0 m/s

a = -4 m/s²

Using 3rd equation of motion,

v² - u² = 2as

0² - (20)² = 2 × -4 × s₂

- 400 = - 8 × s₂

s₂ = 50 m

Total distance (s) = s₁ + s₂

s = 8 + 50

s = 58 m

∴ Total distance travelled by the car = 58 m.

(Diagram has been attached.)

Answer 2

Answer:

Explanation:

Given :-

Initial velocity, u = 20 m/s

Retardation, a = 4 m/s²

Time taken, t = - 0.4 seconds

Final velocity, v = 0 (As car stops)

To Find :-

Distance traveled, s = ??

Formula to be used :-

3rd equation of motion, i.e v² - u² = 2as

Solution :-

s₁ = ut = 20 × 0.4 = 8 m

Putting all the values, we get

v² - u² = 2as

⇒ (0) - (20)² = 2 × (- 4) × s₂

⇒ 400 = 8s₂

⇒ 400/8 = s₂

⇒ s₂ = 50 m

Now, total distance,

⇒ Total distance = s₁ + s₂

⇒ Total distance = 8 + 50

⇒ Total distance = 58 m.

Hence, total distance traveled by car 58 m.

Important terms :-

Distance - Distance is the actual length of the path covered by a body during the whole journey, without taking into consideration its direction. It is a scalar quantity.

Displacement - Displacement is the distance in a particular direction. It is a vector quantity.

Initial speed - Initial speed is that speed with which a body starts its motion in the beginning. It is denoted by u.

Final speed - Final speed is that speed which is acquired by body after its start. It is denoted by v.

Note* - Both are zero when a body finally come to rest or starts from rest.

Retardation - When the final speed of a body is less that its initial speed, the body is said to be retardation. Retardation can also be expressed as acceleration with minus (-) sign.