Question

7. 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?
O
(a) 8100 m
O
(b) 625 m
O
(c) 620 m​

Answer 1

Solution :

⏭ Given:

✏ Initial speed of car = 90kmph

✏ Final speed of car = 0

✏ Acceleration produced in car = 0.5 $\sf{ms^{-2}}$

⏭ To Find:

✏ Distance covered by the car before it is brought to rest.

⏭ Concept:

✏ This question is completely based on concept of stopping distance.

⏭ Formula derivation:

✏ As per second equation of kinematics

$\mapsto \sf \: {v}^{2} - {u}^{2} = 2aS \\ \\ \mapsto \sf \: {0}^{2} - { u }^{2} = 2( - a)S \\ \\ \mapsto \sf \: - {u}^{2} = - 2aS \\ \\ \mapsto \underline{\boxed{ \bold{ \sf{ \pink{S = \frac{ {u}^{2} }{2a}}}}}} \: \star$

✏ Negative sign shows retardation.

⏭ Terms indication:

✏ S denotes distance covered by body

✏ u denotes initial velocity

✏ v denotes final velocity

✏ a denotes acceleration

⏭ Conversation:

✏ 90 kmph = 25 mps

⏭ Calculation:

$\implies \sf \: S = \frac{ ({25)}^{2} }{2 \times 0.5} \\ \\ \implies \sf \: S = \frac{625}{1} \\ \\ \implies \: \underline{ \boxed{ \bold{ \sf{ \purple{S = 625 \: m}}}}} \: \orange{ \bigstar}$

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.