Question

a race car accleration on a straight road from rest to a speed of 180kmhin 25s. assuming uniform accleration of the car throughtout,find the distance coveredin this time.

Answer 1

Answer:

625m

Explanation:

v=180*5/18m/s=50m/s

u=0

t=25s

First apply v=u+at

50=0+a(25)

a=2m/s^2

Now apply S=ut+1/2at^2

S=0(25)+1/2*2*(25)^2

S=625m

Hope it helps

Answer 2

Answer:

$\sf{The \ distance \ covered \ is \ 625 \ m}$

Given:

• Initial velocity (u)=0...[as car was in rest]

• Final velocity (v)= 180 km/hr

• Time=25 s

To find:

• The distance cover by the car in till it's final velocity is 180 km/hr

Solution:

Final velocity (v)=180 km/hr

>> 1 km=1000 m and 1 hr=3600 second

$\therefore$Final velocity (v)=180000/3600=50 m/s

$\boxed{\sf{a=\dfrac{v-u}{t}}}$

$\sf{\therefore{Acceleration =\dfrac{50-0}{25}}}$

$\therefore$ Acceleration=2 m/s²

According to the third equation of Motion.

$\boxed{\sf{v^{2}=u^{2}+2as}}$

$\sf{50^{2}=0^{2}+2(2)(s)}$

$\sf{\therefore{2500=4\times \ s}}$

$\sf{\therefore{s=\dfrac{2500}{4}}}$

$\sf{\therefore{s=625 \ m}}$

$\sf\purple{\tt{\therefore{The \ distance \ covered \ is \ 625 \ m}}}$