Physics, asked by neethu724, 1 year ago

velocity of a car increase for 30 kilometre per hour to 45km/h in 5 minute acceleration to be uniform,calculate the
distance travelled ​

Answers

Answered by deepsen640
37

Answer:

Distance travelled = 3.125 km

Step by step explanations :

given that,

velocity of a car increase for 30 kilometre per hour to 45km/h in 5 minute

here,

initial velocity of the car = 30 km/h

it accelerated uniformly and

gives the

final velocity of the car = 45 km/h

time taken by the car to accelerate uniformly = 5 minutes

changing 5 minutes to hour

= 5/60 h

= 1/12 h

now,

we have,

initial velocity(u) = 30 km/h

final velocity(v) = 45 km/h

time taken(t) = 1/12 h

so,

by the equation of motion,

v = u + at

where,

a is acceleration produced in the car

putting the values,

45 = 30 + a(1/12)

a/12 = 45 - 30

a/12 = 15

a = 180 km/h²

now ,

we have,

initial velocity(u) = 30 km/h

final velocity(v) = 45 km/h

acceleration(a) = 180 km/h²

by the equation of motion,

v² = u² + 2as

where,

S is distance travelled during acceleration

putting the values,

(45)² = (30)² + 2(180)s

360s + 900 = 2025

360s = 2025 - 900

360s = 1125

S = 1125/360

= 3.125 km

so,

Distance travelled = 3.125 km

Answered by ILLIgalAttitude
30

Answer:

Distance travelled = 3.125 km

Step by step explanations:

given,

initial velocity of a car = 30 km/h

final velocity of car = 45 km/h

time taken = 5 minutes

= 5/60 h

= 1/12 h

now,

v = u + at

45 = 30 + a(1/12)

a/12 = 45 - 30

a/12 = 15

a = 180 km/h²

so,

acceleration(a) = 180 km/h²

v² = u² + 2as

S = distance travelled

45² = 30² + 2(180)s

360s = 45² -30²

360s = 2025 - 900

S = 1125/360

= 3.125 km

so,

Distance travelled = 3.125 km

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