Question

a car accelerates uniformly from 18 km/h to 36 km/h in 5s. calculate (1) the acceleration (2) the distance covered by the car in that time

Answer 1

Here Your Answer ✍️

Intial Velocity (u)=18km/h
18 km/h in m/s=18×5/18=5 m/s
Intial Velocity=5m/s

Final Velocity (v)=36 km/h
36 km/h in m/s=36×5/18=10 m/s
Final Velocity=10 m/s

Time(t)=5 seconds

Accerlation (a)=??

Using Equation
V=U+AT
10=5+A×5
5A=10-5
5A=5
A=5/5
A=1 m/s

(1).Accerlation=1 m/s

(2)For Distance Use equation
(v+u/2)×t
(10+5/2)×5
=15/2×5
=37.5 METRES

Distance Covered=37.5 metres

Answer 2

Hey mate

given,

initial velocity of the car U= 18km/h

and Final velocity V= 36km/h

time = 5s

convert 18km/h into m/s= 18×5/18=5m/s

and now 36km/h= 36×5/18= 10m/s

acceleration= final velocity- initial velocity/time taken= 10-5/5= 1m/s
distance covered by car = (v+u/2)×t

{10+5/2}×5= 37.5m

hope it help you