Question

A bus accelerates uniformly from 54 km/h to 72 km/h in 10 seconds Calculate
(i) acceleration in m/s2
(ii) distance covered by the bus in metres during this interval.

Answer 1

(i) Initial velocity, u = 72 km/h

= 72×(5/18) m/s

= 20 m/s

Final velocity, v = 54 km/h

= 54×(5/18) m/s

= 15 m/s

Time in which velocity changes is, t = 10 s

Acceleration, a = (v - u)/t

=> a = (15 - 20)/10

= -0.5 m/s^2

(ii) distance travelled by the bus,

Its avg speed over the 10 sec is (15 + 20)/2

= 17.5 m/sec

17.5*10 = 175 meters

Answer 2

⇒ Given:

A bus accelerates uniformly from 54 km/h to 72 km/h in 10 seconds.

⇒ To Find:

The acceleration of the bus in m/s.

The distance covered by the bus in m during this interval.

⇒ Formulae to be used:

→ $\boxed{\sf{Km/h\:to\:m/s=Multiply\:by\:\dfrac{5}{18}}}$

→ $\boxed{\sf{Acceleration=\dfrac{Rate\:of\:change\:of\:velocity}{Time}}}$

→ $\boxed{\sf{v^2-u^2=2as}}$

⇒ Solution:

At first, we need to convert the given velocities from km/h to m/s. This can be done by multiplying the velocities with 5/18.

Initial velocity [u] = 54 km/h

To convert to m/s:

$\sf{54\times\dfrac{5}{18}=15\:m/s}$

Final velocity [v] = 72 km/h

To convert m/s:

$\sf{72\times\dfrac{5}{18}=20\:m/s}$

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(i) Acceleration in m/s

We know that:

$\sf{Acceleration=\dfrac{Rate\:of\:change\:of\:velocity}{Time}}$

$\sf{Acceleration=\dfrac{v-u}{t}}$

Here,

u= 15 m/s

v = 20 m/s

t = 10 s

Therefore:

$\sf{Acceleration=\dfrac{20-15}{10}}$

$\sf{Acceleration=\dfrac{5}{10}}$

Acceleration = 0.5 m/s²

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(ii) Distance covered by the bus in this interval [in m]

Here, we have to apply the third equation of motion:

v² - u² = 2as

where we know the values of v, u and a.

Applying the values we know in the equation:

$\sf{20^2-15^2=2\times0.5\times\:s}$

$\sf{400-225=1\times\:s}$

$\sf{175=s}$

Distance covered = 175 m

Hence the acceleration of the bus is 0.5 m/s² and the distance covered by the bus is 175 m.