Question

.
A car has initial velocity of 72 km h-1. It is accelerated at 2 ms-2. Calculate the final velocity and the distance coverey
after 3 seconds.

[Ans. 26 m s-1,69 m).​

Answer 1

Answer:

initial velocity ,u=72km/hr.

=72*5/18 m/s

u = 20m/s

accelaration=2m/s^2

time=3 seconds

distance=?

final velocity ,v=?

from first equation of motion :-

v=u+at

v=20+2*3

v=20+6

v=26m/s

from third equation of motion :-

v^2=u^2+2as

26^2=20^2+2*2*s

676=400*4s

676-400=4s

276=4s

s=69metres

Answer 2

Answer:

Final velocity ( v )  = 26 m / sec &  distance ( s ) =  69 m .

Explanation:

Given :

Initial velocity ( u ) = 72 km / hr

For m / sec multiply it by 5 / 18 we get u = 4 × 5 = 20 m / sec.

Acceleration ( a ) = 2 $m/sec^2$

Time ( t ) = 3 sec

From first equation of motion we have

v = u + a t

where v  = final velocity

u = Initial velocity  

a = Acceleration

t = Time

Now put the value in equation we get

v = 20 + 2 × 3 m / sec

v = 20 + 6 m / sec

v = 26 m / sec.

Now we have third equation of motion

$\displaystyle v^2=u^2+2as$

Put the values in equation we get

$\displaystyle (26)^2=(20)^2+2\times2\times s\\\\\displaystyle 676=400+4 s\\\\\displaystyle 676-400=4 s\\\\\displaystyle 4s=276\\\\\displaystyle s=69 \ m$

Thus we get final velocity is 26 m / sec and distance is 69 m .