Question

A car starts from rest and moves with constant
acceleration of 4 m/s2 for 30 seconds. Then the
brakes are applied and the car comes to rest in
another 60 seconds. Find

Ö) Maximum velocity attained by car
(i) Magnitude of retardation of car
(iii) Total distance covered by car
(iv) Average speed of car through out the motion​

Answer 1

Explanation:

so for 30 sec car undergone an acceleration of

4m/s² from rest meabs u=0

now,

a=v-u/t

4=v-0/30==>v=120m/s

then car comes to rest in 60 sec

so at that instance v=0

then v in first case will be u in here

so u=120m/s because for 30 sec it was travelling at 120 m/s so after travelling 120m/s for 30 sec

brakes are applied and the car comes to rest in 60 sec so u=120m/s

a=v-u/t ==> 0-120/60= -20m/s²

if it was asked acceleration then only it will be

-20m/s since retardation is asked so it is negative

acceleration so negative sign is not needed

so retardation is 20m/s

so in first instance

u=0,a=4,t=30

s=ut+1/2at²

s=0×30+1/2×4×900=1800m

in second instance

u=120,t=60,a= -20

s=120×60+1/2×20×3600=43200m since

distance is asked so we can avoid -ve sign

total distance =45000m

average speed = total distance /total time

=45000/60+30=45000/90=500m