Question

if a car
increases its velocity from
18 km/h
to 72km/h in 5 seconds
n
calculate
(i) the acceleration of the car
and Distance covered by
the car 5 seconds​

Answer 1

Answer:

Acceleration= 10.8

distance= 270m

Explanation:

a=v-u/t

a=72-18/5

a=10.8m/s2

now distance= by using third equation of motion: s=u+(1/2)at2

s= 270m

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Answer 2

Answer:

Explanation:

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

v = 72 km/h

u = 18 km/h

t = 5 secs

$a = \frac{72-18}{5}$

$a = \frac{54}{5} =10.8$  $m/s^2$

a = 10.8 $m/s^2$

Using 3rd equation of motion,

$s=ut+\frac{at^2}{2}$

u = 18 km/h

t = 5 secs

a = 10.8 $m/s^2$

$s = 90+135$

s = 225 m

Hope it helps...