Question

A bus accelerates uniformly from 72km/h to 108km/h in 10s (a) find acceleration (b) find the distance covered by the bus in 10s.​

Answer 1

Answer:

given

u=72km/h=72*5/18=20m/s

v=108km/h=108*5/18=30m/s

t=10s

now from first eq of motion

v=u+at

30=20+10a

a=1m/s^2

now

2nd eq of motion

s=ut + 1/2at^2

s=20*10+1/2*1*10*10

s=200+50

s=250m

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Answer 2

Answer:

(a)The acceleration of bus is $1m/s^2$

(b)The distance traveled by bus is 250m.

Explanation:

Given data is,

Initial velocity of bus is

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

Final velocity of the bus is

$v=108km/h=108\times\frac{5}{18}m/s=30m/s$

Time taken is $t=10sec$

(a)The acceleration is given by the relation

$a=\frac{v-u}{t}$

Substituting the given values,we get

$a=\frac{30-20}{10}=1m/s^2$

(b)The distance covered is given by the relation

$v^{2}-u^2=2as= > s=\frac{v^2-u^{2}}{2a}$

Substituting the given values,we get

$s=\frac{30^2-20^2}{2(1)}=250m$

Therefore,

The acceleration and distance covered by bus are $1m/s^2$ and $250m$ respectively.

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