Question

A car accelerates uniformly from 18km per hour to 36km per hour in 5 seconds . Calculate the acceleration and the distance covered by the car in that time.

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Answer 1

Answer:

Given,

u=18km/h=

60×60

18×1000

=5m/s

v=36km/h=

60×60

36×1000

=10m/s

t=5sec

acceleration, a=?

1st equation of motion,

v=u+at

10=5+5a

5a=5

a=

5

5

=1m/s

Explanation:

Hope this answer will be helpful..

Answer 2

Given :-

A car accelerates uniformly from 18km per hour to 36km per hour in 5 seconds.

To Find :-

The acceleration and the distance covered by the car in that time.

Solution :-

We know that,

• u = Initial velocity
• t = Time
• v = Final velocity
• a = Acceleration

Initial velocity (u) =18 km/hr

By converting,

$\sf u=\dfrac{18 \times 1000}{60 \times 60}$

$\sf u=5 \ m/s$

Final velocity (v) = 36 km/hr

By converting,

$\sf \sf v=\dfrac{36 \times 1000}{60 \times 60}$

$\sf v=10 \ m/s$

By the formula,

$\underline{\boxed{\sf Acceleration=\dfrac{(v-u)}{t} }}$

Substituting their values,

$\sf a=\dfrac{(10-5)}{5}$

$\sf a=\dfrac{-5}{5}$

$\sf a = 1 \ m/s^2$

Therefore, the acceleration is 1 m/s².

By the formula,

$\underline{\boxed{\sf Distance \ travelled =ut+\dfrac{1}{2} a \times t^2 }}$

Substituting their values,

$\sf =5 \times 5+\dfrac{1}{2} \times 1 \times 5^2$

$\sf =25+\dfrac{1}{2} \times 25$

$\sf = 25 + 12.5$

$\sf = 37.5 \ m$

Therefore, the car covers 37.5 m at that time.