Question

A car is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of – 0.5 m/s2. Find how far the car will go before it is brought to rest?

Answer 1

A car is travelling at a speed of 90 km/h (initial velocity of the car is 90 km/hr). Brakes are applied (means final velocity of the car is 0 m/s) so as to produce a uniform acceleration of -0.5 m/s² (negative sign shows retardation).

We have to find the distance covered by the car.

First equation of motion:

v = u + at

Second equation of motion:

s = ut + 1/2 at²

Third equation of motion:

v² - u² = 2as

From above data we have; u is 90 m/s, v is 0 m/s and a is -0.5 m/s². So, using the Third Equation Of Motion i.e. v² - u² = 2as

Firstly convert km/hr into m/s. To do this, multiply the given value by 5/18.

→ u = 90*5/18 = 25 m/s

Substitute the known values in the above formula,

→ (0)² - (25)² = 2(-0.5)s

→ 0 - 625 = - 1s

→ 625 = s

Therefore, the distance covered by the car is 625 m.

Answer 2

Hey Pretty Stranger!

Here, we've :

Initial Velocity (u) = 90 km/hr = 90 × 5/18 = 25 m/s

Final Velocity (v) = 0 m/s

Acceleration (a) = -0.5 m/s²

We can solve this question either by finding time first and then using 2nd equation of motion or directly using 3rd equation of motion.

Here, I'm solving it using 3rd equation of motion

$\dag \: \: \large \boxed{ \sf \: {v}^{2} = {u}^{2} + 2as}$

$\longrightarrow \sf \: ( {0})^{2} = (25)^{2} + 2( - 0.5)(s)$

$\longrightarrow \sf \: 0= 625 - s$

$\longrightarrow \sf \: 0 + 625 = s$

$\longrightarrow \sf s = 625 \: m$

$\therefore$ The car will go 625 m before it is brought to rest.

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The other two equations of motions are :

$\sf \: i) \: v = u + at$

$\sf \: ii) \: s = ut \: + \dfrac{1}{2} a {t}^{2}$

Note : Symbols have their usual meaning.