Question

28. A car starts from rest and accelerates uniformly for 10 s to a velocity of 8 m/s. Then it runs at a constant velocity and is finally brought to rest in 64 m with a constant retardation. The total distance covered by the car is 584 m. Find the value of acceleration, retardation and total time taken.​

Answer 1

V=at

⇒8=a10

⇒a=0.8m/s2

V2=U2+2as

⇒U=82+2×a64

⇒a=2×64−64=−0.5m/s2

Distance travelled= Area of graph

⇒584=21×10×8+8×t1+21×16×8

⇒584=40+8t1+64

⇒8t1=480

⇒t1=60

Total time= 10+60+16=86s

Answer 2

Given,

Time = 10 s

Final velocity = 8 m/s

Initial velocity = 0 m/s

Distance travelled = 64 m

Total distance = 584 m

Acceleration = v + u / t

Acceleration = 8 + 0/ 10

Acceleration = 8 /10

Acceleration = 0.8 m/ s²

v² = u² + 2as

0 = (8)² + 2 × a × 64

0 = 64 × a × 128

- 64 = a × 128

- 64/128 = a

- 0.5 = retardation

Distance travelled

s = ut × 1/2 × at²

= 0 × 10 + 1/2 × 0.8 × (10)²

= 0 + 1/2 × 8/10 ×10×10

= 40 m

Total distance travelled =

= 40 + u + 64 = 584

= 104 + u = 584

= u = 584 - 104

= u = 480 m

t1 =

d = v × t

480 = 8 × t

480 / 8 = t

t = 60s

t2 =

= retardation / final velocity

= 0.5 / 8

= 16s

Total time taken = 10s + 60s + 16s

= 86 s