Question

a car move with uniform velocity of 40 ms-1 for 5 s it comes to rest in the next 10 s with uniform decelration find decelration total distance travelled by a car

Answer 1

Given:-

• Initial velocity,u = 40m/s

• Final velocity ,v = 0m/s

• Total time taken ,t = 10+5 = 15s

To Find:-

• Uniform deceleration ,a
• distance covered ,s

Solution:-

By using 1st equation of motion

→ v = u+at

Substitute the value we get

→ 0 = 40 + a×15

→ - 40 = a×15

→ a = -40/15

→ a = -2.66 m/s²

∴ The deceleration of the car is 2.66m/s²

Now , Using 3rd equation of motion

• v² = u² +2as

Substitute the value we get

→ 0² = 40² + 2× (-2.66) × s

→ 0 = 1600 + (-5. 32) ×s

→ -1600 = -5.32 ×s

→ s = -1600/-5.32

→ s = 300.75 m

∴ The distance covered by the car is 300.75 Metres.

Answer 2

Answer:

A car moves with uniform velocity (Initial Velocity, u) of 40 m/s for 5 seconds it comes to rest in the next 10 seconds. Therefore, Total time taken (t) is 10 + 5 = 15 seconds and the Final Velocity (v) of car is 0 m/s.

We are supposed to calculate the Uniform decelration (a) and distance covered (s) by car.

We have 3 equations of motion, let's see what are the 3 equations of motion :

• v = u + at
• s = ut + ½ at²
• v² = u² + 2as

Now, just use the suitable equation on based on given data. So, from given data we understand that here we can use first equation of motion :

Lets find the decelration (a) :-

$=> \sf v = u + at$

$=> \sf 0 = 44 + a(15)$

$=> \sf - 40 = 15a$

$=> \sf a =\dfrac{ - 40}{15}$

$=> \sf a =2.66 \: m/s^2$

Now, we can use third equation of motion to calculate distance. travelled by car :

$=> \sf {v}^{2} = {u}^{2} + 2as$

$=> \sf {0}^{2} = {40}^{2} + 2( - 2.66)s$

$=> \sf - 1600 = - 5 .32 \times s$

$=> \sf s = \dfrac{ - 1600}{ - 5.32}$

$=> \sf s = \dfrac{1600}{ 5.32}$

$=> \sf s = 300.75 \: m$

⛬ The decelration of car is 2.66 m/s² and distance coverd by the car is 300.75 meters.