Question

A motar car start from rest and acceleration uniformly for 10sec to a velocity 20ms-1 . it than runs at a constant speed and is finally brought to rest in 40meter with a constant acceleration total distance cover in 640 metre. find the value of acceleration, retardation and total time taken​

Answer 1

Answer:

V=at

⇒8=a10

⇒a=0.8m/s

2

V

2

=U

2

+2as

⇒U=8

2

+2×a64

⇒a=

2×64

−64

=−0.5m/s

2

Distance travelled= Area of graph

⇒584=

2

1

×10×8+8×t

1

+

2

1

×16×8

⇒584=40+8t

1

+64

⇒8t

1

=480

⇒t

1

=60

Total time= 10+60+16=86s

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Answer 2

initial velocity = 0 m/s

final velocity in case 1 = 20 m/s

time in case 1 = 10 s

acceleration in case 1 = (v-u)/t

=> (20-0)/10

=> 20/10

=> 2 m/s²

distance = 640 m

It is given that the speed is constant which is 20 m/s

initial velocity in case 2 = 20 m/s

final velocity in case 2 = 0 m/s [ as brought to rest ]

acceleration = 0 m/s² [ as velocity is constant ]

We will find time through 2nd equation of motion.

2nd equation of motion :-

$s = ut + \frac{1}{2}a {t}^{2}$

640 = 20t + (1/2 × 0 × t²)

640 = 20t + 0

20t + 0 = 640 ( transposing )

20t = 640

t = 640/20

t = 32

therefore, the time is 32 seconds and the retardation is 0 m/s²