Question

A car accelerates uniformly from 18 km h to 36km h in 5 sec calculate acceleration and distance cover by car in that time

Answer 1

Acceleration = Change in Velocity/Time
Change in Velocity = 36-18 = 18 km/h=5 m/s
Time= 5 Seconds
Acceleration = 5/5= 1 m/s2

Equation of motion,s=ut+(1/2)at2
u=18 km/h=5 m/s
t=5 s
a=1 m/s2
s= (5*5)+(1/2*1*5*5)
s=25+12.5 i.e., s=37.5 m
Hope you are clear with my explanations

Answer 2

Answer:

Explanation:

Solution,

Here, We have initial and final velocity and time taken.

Initial velocity, u is given as 18 km/h, So we have to convert it into m/s,

For that we have to multiply it with 5/18,

So, 18 × 5/18 = 5 m/s.

Similarly, the final velocity is given as 36 km/s.

So, 36 × 5/18 = 10 m/s.

Now, time taken is also given as 5 seconds,

Here, we have to find,

Acceleration and Distance covered.

According to the first equation of motion,

We know that,

v = u + at

So, putting all the values, we get

⇒ v = u + at

⇒ 10 = 5 + a × 5

⇒ 10 - 5 = 5a

⇒ 5 = 5a

⇒ 5/5 = a

⇒ a = 1 m/s².

Hence, the acceleration of the car is 1 m/s².

Now, we have to find the distance covered, s

According to the 3rd equation of motion,

we know that,

v² - u² = 2as

So, putting all the values again, we get

⇒ v² - u² = 2as

⇒ (10)² - (5)² = 2 × 1 × s

⇒ 100 - 25 = 2s

⇒ 75 = 2s

⇒ 75/2 = s

⇒ s = 37.5 m.

Hence, the distance covered by car is 37.5 m.