Question

starting from rest , a formula one car accelerates uniformly at 25m\s2 for 30secs. what distance does it cover in the last one second of motion​

Answer 1

The distance covered in the last second of motion is 737.5 m

Explanation:

The motion of the car is a uniformly accelerated motion, so we can use the suvat equations.

First of all, we have to find the velocity of the car when the last second of motion starts, that is the velocity of the car after t = 29 s. We can use the equation:

v = u + at

where

u = 0 is the initial velocity

$a=25 m/s^2$ is the acceleration

Substituting t = 29 s,

$v=0+(25)(29)=725 m/s$

Now we can find the distance covered in the last second of motion by using

$s=ut+\frac{1}{2}at^2$

where

u = 725 m/s is the velocity at the beginning of the last second

t = 1 s is the time interval considered

$a=25 m/s^2$ is the acceleration

Substituting,

$s=(725)(1)+\frac{1}{2}(25)(1)^2=737.5 m$

Note that the acceleration of $25 m/s^2$ is not realistic for a car, but I still have used the data of the problem.

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