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

Answer:

So the car displacement after 3.7 sec is 0.030 km

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 $s=ut+\frac{1}{2}at^2$, here s is displacement, u is initial velocity, t is time , and a is acceleration

So displacement $s=ut+\frac{1}{2}at^2=6.64\times 3.7+\frac{1}{2}\times 0.85\times 3.7^2=30.386m$

We know that 1 km = 1000 m

So 30.386 m = 0.030 km

Answer 2

Answer:

The  displacement of car after that time is 30.56 m.

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

$v^2=u^2+2as$

$s=\dfrac{v^2-u^2}{2a}$

Put the value into the formula

$s=\dfrac{9.8^2-6.64^2}{2\times0.85}$

$s =30.56\ m$

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