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15 days ago

A bus slows down uniformly from 75.0 km/h to 0 km/h in 21 s. How far does it travel before stopping?

2 answers:

7 0

1 hour = 3,600 seconds
1 km = 1,000 meters

75 km/hour = (75,000/3,600) m/s = 20-5/6 m/s

The mean speed during the deceleration is

(1/2)(20-5/6 + 0) = 10-5/12 m/s.

Traveling at this average speed for 21 seconds,
the bus covers

(10-5/12) × (21) = 218.75 meters.

kicyunya [1K]15 days ago

4 0

According to the problem, the bus decelerates uniformly from 75 km/h to 0 km/h.

Thus the initial speed u = 75 km/h and the final speed v = 0 km/h.

The time to stop is t = 21 s.

Convert units: 1 km = 1000 m and 1 hour = 3600 s.

Therefore u = $75 km/h =75*\frac{1000 m}{3600s}$ = 20.83 m/s.

We need the stopping distance.

From kinematics, v = u + a t, where a is the acceleration.

With v = 0, u = 20.83 m/s, and t = 21 s,

0 = 20.83 + a × 21

so a × 21 = -20.83

hence a = -[20.83] ÷ 21 = $= -0.9919 m/s^2$ (the negative sign shows deceleration)

Using the displacement formula $s = ut +\frac{1}{2} at^2$ and substituting the known values gives

$s = 20.83 *21 -\frac{1}{2} 0.9919 *[21^2]$

which results in s = 218.7 metre. [ans]

Therefore the bus stops after 218.7 m.

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