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/s^2. Find how far the car will go before it is brought to rest?
(a) 8100 m (b) 900 m (c) 625 m (d) 620 m

Answer 1

Given :-

Initial velocity of car (u) =90km/h

Convert in m/s

$\sf \cancel{90}\times \dfrac{5}{\cancel{18}} = 5\times 5\\ \\ \sf \ \ u = 25m/s$

Final velocity (v) = 0 [ breaks applied ]

Acceleration (a) = -0.5 m/s^2

To find :-

The distance covered by car before it come in rest

$\underline{\sf{\dag\ \ By \ 3rd \ equation\ of \ Motion }}$

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

• Where

• v= Final velocity

•u= Initial velocity

•a= acceleration

•s= Distance

• Now calculate the distance

$\implies\sf v^2=u^2+2as\\ \\ \implies\sf v(0)^2= (25)^2\times 2\times (-0.5)\times s\\ \\ \implies\sf 0= 625 - 1s\\ \\ \implies\sf 0-625= -s \\ \\ \implies\sf \cancel{-}625= \cancel{-} s\\ \\ \implies\sf 625= s\\ \\ \implies {\boxed{\sf { s= 625m}}}$

So option 3 is correct

$\underline{\boxed{\textbf{\textsf{ Distance \ covered \ By \ car \ before \ stop = 625m}}}}$

Answer 2

Answer:

Explanation:

Given:-

Initial speed of the train, u = 90 km/h = 25 m/s

The final speed of the train, v = 0

Acceleration = - 0.5 m/s²

To Find:-

Distance acquired

Formula to be used:-

Third equation of motion:

v² = u² + 2 as

Solution:-

Putting all the value, we get

v² = u² + 2as

⇒ (0)² = (25)² + 2 (- 0.5) s

⇒ s = 625 m

Hence, The train will cover a distance of 625 m before coming to rest.