Question

A car with an initial speed of 6.64 m/s accelerates at a uniform rate of 0.85 m/s^2 for 3.7s. The final speed of the car is 9.8 m/s. What is the cars displacement after that time? answer in km.

Answer 1

The displacement of car after that time is 30.56 m.

Step-by-step explanation:

Given that,

Initial velocity = 6.64 m/s

Acceleration = 0.85 m/s²

Time = 3.7 s

Final velocity = 9.8 m/s

We need to calculate the displacement

Using equation of motion

`v^2=u^2+2as`

`s=(v^2-u^2)/(2a)`

Put the value into the formula

`s=(9.8^2-6.64^2)/(2*0.85)`

`s =30.56\ m`

Hence, The displacement of car after that time is 30.56 m.

Answer 2

So the car displacement after 3.7 sec is 0.030 km

Step-by-step explanation:

We have given initial velocity u = 6.64 m/sec

Acceleration
`a=0.85m/sec^2`

Time t = 3.7 sec

Final velocity v = 9.8 m/sec

We have to find the displacement after that time

From second equation of motion we know that
`s=ut+(1)/(2)at^2`, here s is displacement, u is initial velocity, t is time , and a is acceleration

So displacement
`s=ut+(1)/(2)at^2=6.64* 3.7+(1)/(2)* 0.85* 3.7^2=30.386m`

We know that 1 km = 1000 m

So 30.386 m = 0.030 km