Question

A car starts from rest and acquire on velocity of 54km/hr in 2min.find acceleration.and distance travelled

Answer 1

✬ Distance = 900 m ✬

✬ Acceleration = 0.125 m/s² ✬

Explanation:

Given:

• A car tarted from rest i.e Initial velocity is 0.
• Final velocity of car is 54 km/hr
• Time taken is 2 minutes.

To Find:

• What is the acceleration and Distance covered by car ?

Formula to be used:

• v = u + at (for acceleration)
• s = ut + 1/2at² ( for distance )

Solution: First change the units in m/s to make units same.

➯ 1 minutes = 60 seconds

➯ 2 minutes = 2(60) = 120 seconds.

Multiply with 5/18 to change km/h to m/s.

➯ Final velocity = 54(5/18)

➯ v = 15 m/s.

Now, put the values on formula.

$\implies{\rm }$ v = u + at

$\implies{\rm }$ 15 = 0 + a $\times$ 120

$\implies{\rm }$ 15 = 120a

$\implies{\rm }$ 15/120 = a

$\implies{\rm }$ 0.125 m/s² = a

So, the acceleration of car is 0.125 m/s².

__________________________

Using second formula for finding distance

$\implies{\rm }$ s = ut + 1/2at²

$\implies{\rm }$ s = 0 $\times$ 120 + 1/2 $\times$ 0.125 $\times$ 120²

$\implies{\rm }$ s = 0.125/2 $\times$ 14400

$\implies{\rm }$ s = 0.125 $\times$ 7200

$\implies{\rm }$ s = 900 m

Hence, the distance covered by car is 900 m.

Answer 2

Answer:

1620 km/h and 0.9 km

Explanation:

here the initial velocity(u) = 0

and velocity after 2 min (v) = 54 km/h

we have newton equation v = u + at

54 km/h= 0 + a 0.033 h

here a is acceleration

a = 1620 km/h^2

now we have second newton equation

s = ut + 1/2 at^2

here u is 0

∴s = 1/2 a t^2

here s is distance

s = 0.9 km