Question

A car accelerates uniformly from 18 km/h to
36km/h in 5s calculate
a) acceleration
b) distance covered by the cay in that came time.​

Answer 1

Answer:

a)1m/s2 , b) 37.5m

Explanation:

Given , initial velocity (u) =18km/h = 18*5/18

=5m/s

final velocity (v) = 36km/hr = 36*5/18

=10m/s

time (t) = 5s

a) acceleration = Change in velocity /time

10-5/5= 1m/s^2.

b) distance

Using 2nd equation of motion:

s=ut+1/2at^2

Putting the values-

s= 5*5+1/2*1*5^2

25+12.5

37.5m.Ans

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Answer 2

Given -

Initially velocity, u = 18 km/hr

⠀⠀⠀⠀⠀⠀ ⠀⠀⠀ = 5 m/s

Final velocity ,v = 36 km/hr

⠀⠀⠀⠀⠀⠀ ⠀⠀= 10 m/s

Time taken,t = 5 sec

To find,

• Acceleration
• Distance covered

Solution-

Acceleration, a = $\large \frac{v - u}{t}$

$\frac{(10 \: m/s - 5 m/s)}{5s}$

$acceleration \: = \: 10 \: m/s^{2}$

Distance covered-

By using 2nd equation of motion we will solve it.

$s = ut + \frac{1}{2}a {t}^{2}$

$s = 5m/s \times 5 s + \frac{1}{2} \times 10 m/s \times ( {5})^{2}$

$s = 25 + 125$(as 2 will get cancelled with 10 leaves 5 then 5 will get multipled by 25)

$s = 150 m$

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