Question

a car starts from rest and accelerates uniformly to 40km/hr in 27s. It then continues to move at uniform speed for the next 3 min. The breaks are then applied to bring the car at the rest in 18s.calculate the total distance travelled by the cars.​

Answer 1

Answer:

2250 m

Explanation:

Car is at rest(u = 0), time taken is 27s.

v = 40 kmph = 100/9 m/s

Acceleration = (v - u)/t = 100/243 m/s²

Using equation of motion:

S = ut + ½ at² = 0 + ½ (100/243) (27)²

S = ½ (100 x 9) = 150 m

It then moves with uniform speed, *no acceleration(a = 0), last speed = 100/9, time in this period = 3 min = 180 s.

S = ut + ½ at² = (100/9)(180) + ½(0)t²

S = 2000 m

Speed of the car is still same(u = 100/9), final speed(v = 0, as it comes to rest in t = 18s).

de-acceleration = (v - u)/t = - 50/81 m/s²

S = ut + ½ at²

S = (100/9)(18) + ½ (-50/81) (18)²

S = 200 - 100 = 100 m

Total distance = 150m + 2000m + 100m

Total distance = 2250 m

Answer 2

Given :-

A car starts from rest and accelerates uniformly to 40km/hr in 27s. It then continues to move at uniform speed for the next 3 min. The breaks are then applied to bring the car at the rest in 18s.

To Find :-

Distance travelled

Solution :-

In first case

1 km/h = 5/18 m/s

40 km/h = 40 × 5/18 = 200/18 = 100/9 m/s

Now

v = u + at

100/9 = 0 + a(27)

100/9 = 0 + 27a

100/9 = 27a

100/9 × 1/27 = a

100/243 = a

0.41 = a

s = ut + 1/2 at²

s = 0 × 27 + 1/2 × 0.41 × (27)²

s = 0 + 1/2 × 0.41 × 729

s = 0 + 150

s = 150 m

In second case

1 min = 60 sec

3 min = 180 sec

s = ut + 1/2 at²

s = (100/9) × 180 + 1/2 × 0 × (180)²

s = 100 × 20 + 1/2 × 32400

s = 2000 + 32400

s = 19500 m

Now

Decleration = - 50/81 m/s²

s = ut + 1/2 at²

s = 100/9 × 18 + 1/2 × -50/81 × 18²

s = 100 × 2 + 1 × -25/81 × 324

s = 100 m

Now

s = 150 + 19500 + 100

s = 19750 m

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