Question

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Answer 1

Explanation:

u = 0

a = 4 m/s²

t = 30 sec

at = v - u

4*30 = v - 0

120 = v

120 m = v ( highest velocity)

retardation = - a

u = 120 m

v = 0

t = 60

a = ( v - u) / t

a = - 20 m / s² ( magnitude of retardation)

speed * time = distance

120 * 90 = 10600 m

Average speed = total distance / total time

= 10600/ 90

= 120 m/ s²

Answer 2

Answer:

120 m/s, -2 m/s2, 5400m, 60 m/s

Explanation:

In first 30 second car's velocity is increasing and after that it is decreasing. so maximum velocity is at 30 seconds. Call it V30.

V30 = V0 + a x t

V30 = 0 + 4 x 30

V30 = 120 m/s2

Distance covered in 30 seconds,

d30 = [(V30)^2 - (V0)^2]/[2×a]

=[120^2-0]/[2×4]

= 1800 m

Now in next 60 seconds car's velocity comes down from V30= 120 to 0 m/s.

So, using above formula,

0 = V30 + a x 60

0 = 120 + 60a

a= -2 m/s2 . so retardation magnitude = 2 m/s2

distance covered in 60 seconds= d60

= {(V60)^{2} - (V30)^2}/{2*a}

= { 0^2 - 120^2 } / {2*(-2)}

=3600 m

d60= 3600 m

Total distance covered = d30+d60 = 1800+3600 =5400 m

Average speed =

Total distance travelled/Total time

= 5400/90 = 60 m/s