Question

if a bus travelling at 600 km per hour is subjected to a steady declaration of 5 metre per second square how long will it take to come to rest​

Answer 1

Answer:

33.33 seconds

Explanation:

Given:

Initial velocity = u = 600 km/hr

$600 \frac{km}{hr} = 600 \times \frac{5}{18} = \frac{500}{3} \frac{m}{s}$

Acceleration = a = -5 m/s^2

Final velocity = v = 0 (As the bus would finally come to rest)

To find:

Time

Using the first equation of motion which says :

V=u+at

Substituting the values, we get:

$0 = \frac{500}{3} + (- 5)t$

$0 = \frac{500}{3} - 5t$

$- \frac{500}{3} = - 5t$

$t = \frac{ - 500}{3} \times \frac{ - 1}{5}$

$t = \frac{100}{3} = 33.33 \: seconds$

The time taken to stop is 33.33 seconds

Answer 2

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QUESTION :

if a bus travelling at 600 km per hour is subjected to a steady declaration of 5 metre per second square how long will it take to come to rest...

SOLUTION :

The innitial velocity of the bus is 600 kmph

=> 600 × 5 / 18 m / s

=> 166.67 m / s

A = -5 m / s ^ 2

Let the time taken for the bus to come to rest be t

=> Using Newton' first law of motion ,

A = v - u / t

Where , V is the final velocity .

U Is the initial velocity.

T is the time taken.

A is the acceleration.

A = -5 m / s ^ 2

V = 0 as the bus stops after travelling some distance.

So,

-5 = 0 - 166.67 / t

The negative sign on both sides gets cancelled.

Now, t = 166.67 / 5 s = 33.334 seconds approximately.

So , the time taken for the bus to come to rest is 33.334 Seconds approximately.

ANSWER :

THE time taken for the bus to come to rest is 33.334 Seconds approximately.