Question

a uniformly accelerated car changes its velocity from 18km/hr to 36km/hr in 5 second then determine to acceleration and distance covered ​

Answer 1

Given :-

▪ A uniformly accelerated car changes it velocity from u = 18 km/h to v = 36 km/h in 5 seconds.

To Find :-

▪ Acceleration and distance covered.

Solution :-

First, Let us convert the velocities into m/s

The conversion factor is 5 / 18

⇒ u = 18 km/h

⇒ u = 18 × 5/18 m/s

⇒ u = 5 m/s

Also,

⇒ v = 36 km/h

⇒ v = 36 × 5/18 m/s

⇒ v = 2 × 5 m/s

⇒ v = 10 m/s

Now, We have

• u = 5 m/s
• v = 10 m/s
• t = 5 seconds

Using the first equation of motion, acceleration can be calculated as,

⇒ v = u + at

⇒ 10 = 5 + a × 5

⇒ 10 - 5 = 5a

⇒ a = 5/5

⇒ a = 1 m/s²

Now, We have to find the distance covered during this interval. So, we have

• u = 5 m/s
• v = 10 m/s
• a = 1 m/s
• t = 5 seconds

The distance travelled can be calculated using either the second equation or the third one.

Using the third equation of motion,

⇒ 2as = v² - u²

⇒ 2×1×s = (10)² - (5)²

⇒ 2s = 100 - 25

⇒ 2s = 75

⇒ s = 37.5 m

So,

• Acceleration = 1 m/s²
• Distance travelled = 37.5 m

Answer 2

Answer:

accerlation = 1 m/s^2

distance = 37.5m

Explanation:

initial velocity ( in m/s) = 18 * 5/18 = 5 m/s

fimal velocity                = 36 *5/18 = 10m/s

accerlation  = change in velocity / time

                   = 10 - 5 / 5

                   = 1m/s^2

distance by formula

v^2 -u ^2 = 2as

s = v^2 -u ^2 / 2a

s = 100 - 25 / 2 * 1

s = 75 / 2

s = 37.5m