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

Answer:

Initial velocity of bus(u) = 54km/h

= 54 × 5/18

= 15m/s

Final velocity of bus(v) = 72km/h

=72×5/18

= 20m/s

Time taken(t) = 10 seconds

(i) Acceleration (a) = v-u/t

= 20-15/10

= 0.5 m/s^2

(ii) From 2nd equation of motion

s = ut + 1/2at^2

s = 15×10 + 1/2×0.5×10^2

s = 150 + 25

s = 175m

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Answer 2

Answer:

(i) 0.5 m/s²

(ii) 175 metres

Step-by-step explanation:

Given:

• Initial velocity of the bus = u = 54 km/h
• Final velocity of the bus = v = 72 km/h
• Time taken = 10 seconds

To find:

• Acceleration
• Distance covered by the bus

Initial velocity = 54 km/h = $54 \times \frac{5}{18} = 15 \ m/s$

Final velocity = 72 km/h = $72 \times \frac{5}{18} = 20 \ m/s$

(i)

Acceleration = (final velocity - initial veloicty)/time

Acceleration = (20-15)/10

Acceleration = (5)/10

Acceleration = 0.5 m/s²

(ii)

Using third equation of motion:

V²-U²=2as

20²-15²=2×0.5×s

400-225=1s

175=1s

s = 175 metres