Question

Brakes are applied to a car so as to produce a retardation of 5 m/s*2. If the car takes 1.5 s to stop after applying the brakes, calculate the distance travelled by it?

Answer 1

AnswEr :

We are given that a retardation of 5 m/s² i.e Acceleration = -5 m/s² is produced by the car after applying the brakes. The car takes 1.5 second to stop after applying brakes i.e time (t) = 1.5 s. Here the Final Velocity (v) will be 0 m/s. We need to calculate the distance travelled by the car.

• First of all we need to calculate the initial velocity (u) of the car by using first kinematical equation of motion.

• 1ˢᵗ⠀E Q U A T I O N O F M O T I O N :

↠ v = u + at

↠ 0 = u + (-5)(1.5)

↠ 0 = u - 7.5

↠ u = 7.5 m/s

◗ Initial Velocity (u) = 7.5 m/s

________________________

• Now, by using second kinematical equation of motion we can calculate the distance travelled by the car :

•2ⁿᵈ E Q U A T I O N O F⠀M O T I O N :

↠s = ut + ½ at²

↠ s = 7.5 × 1.5 + ½ × (-5) × (1.5)²

↠ s = 11.25 + (⁻⁵/₂) × 2.25

↠ s = 11.25 - 5 × 1.125

↠ s = 11.25 - 5 × 1.125

↠ s = 11.25 - 5.625

↠ s = 5.625 m

◗ Distance travelled (s) = 5.625 m

∴ The distance travelled by the car is 5.625 m .

Answer 2

$\dag\:\underline{\sf Question:-}$

Brakes are applied to a car so as to produce a retardation of 5 m/s². If the car takes 1.5 s to stop after applying the brakes, calculate the distance travelled by it?

$\dag\:\underline{\sf Answer:-}$

$\star \: Given:-\left[\begin{array}{ccc}Acceleration(a) = -5m/s^2\\Time(t) = 1.5 secs\\Final \: velocity = 0 m/s\end{array}$

Acceleration is negative, as we have been provided with retardation/deceleration, which is negative acceleration.

∴ Acceleartion = -5 m/s²

$\star \: To \: find:- Distance(s) \: travelled \: by \: the \: car \: after \: the \: brakes \: were \: applied$

$\star \: Solution:-$

Using first kinematical equation:-

⇒ v = u + at

⇒ (0) = u + (-5) (1.5)

⇒ 0 = u - 7.5

⇒ u = 7.5 m/s

∴ Initial velocity(u) of the car = 7.5 m/s

_______________________________________

Using second equation of motion:-

$s = ut +\frac{1}{2} at^2$

$s = (7.5)(1.5) + \frac{1}{2} (-5)(1.5)^2$

$s = 11.25 + (\frac{-5}{2}) (2.25)\\s = 11.25 - 5 \times 1.125\\s = 11.25 - 5.625$

$\textbf{s = 5.625}$

∴ Distance covered by the car after brakes were applied = 5.625 m

Hope this helps! ♡