Question

A bus was moving with a speed of 54km/h. on applying brakes, it stopped in 8 sec. calculate the acceleration and the distance travelled bus stopping.

Answer 1

Heya!!

Here's your answer!!!





initial velocity (v) = 54km/hr or 15m/sec

Final velocity (u) = 0m/sec

Time(t) = 8 sec

acceleration (a) = ? (to find)

as

$a = \frac{v - u}{t} \\ \\ a = \frac{0 - 15}{8} \\ \\ a = \frac{ - 15}{8} \\ \\ a = - 1.875m \: {sec}^{ - 2}$

there acceleration = -1.875m/ sec×sec.

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@vaibhavhoax
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Answer 2

Answer:

I would do it this way:

We know out initial velocity (u) to be 54km/hr.

This comes to 15 m/s. (54 × 1000/3600) or

(54 ×5/18)

Thus u= 15 m/s.

Time taken to stop = 8 seconds.

Thus t = 8s.

The bus stops completely. Thus final velocity,(v) = 0 m/s.

Now, we can use the equation:

v=u+at

Now v = 0 , u = 15 and t= 8.

=>0=15+8a.

Rearranging,

=>a=−15/8.

Thus acceleration comes to −1.875m/s2.

Now we know acceleration, time and initial velocity. Thus distance can be calculated as:

s=ut+(at2)/2.

=>s=15×8+(−1.875×64)/2.

Solving this, we get the distance to be 60m.

P.S. The acceleration has a negative value because the value of initial velocity is greater than the value of final velocity. As the bus applies brakes.

This is known as deceleration.

Explanation: