Question

Find the acceleration required to bring a 1500-kg car to rest from a speed of 100 km/h

Answer 1

The acceleration of the car is $7.03 m/s^2$

Explanation:

The motion of the car is a uniformly accelerated motion, so we can find its acceleration using the following SUVAT equation:

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance covered

In this problem we have

u = 0 (the car starts from rest)

$v= 100 km/h \cdot \frac{3600 s/h}{1000 m/km}=27.8 m/s$ is the final velocity

s = 55 m is the distance

solving for a,

$a=\frac{v^2-u^2}{2s}=\frac{(27.8)^2-0}{2(55)}=7.03 m/s^2$

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Answer 2

The acceleration required to bring a 1500 kg car to rest is $-7 m s^{-2}$

Explanation:

As the car is bought to rest from the moving speed of 100 km/h, the deceleration exhibited by the car can be calculated provided the distance covered by the car before coming at rest. So the deceleration can be obtained using third equation of motion.

So in the present case, the displacement of the car before it came to rest i.e. s = 55 m, while the initial velocity will be the speed at which the car is moving and final velocity will be zero as the car will stop after travelling for 55 m. So the acceleration will be as below:

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

                       $a=\frac{v^{2}-u^{2}}{2 s}=\frac{0-(100)^{2}(k m / h)^{2}}{2 \times\left(\frac{55}{1000}\right) k m}$

Thus,

                        $a=\frac{-10000 \times 1000}{2 \times 55}=-90909.09 \mathrm{km} / h^{2}$

Converting km to meter and hours to seconds, we get

                        $a=\frac{-90909.09 \times 1000 m}{3600 \times 3600 s^{2}}=-7 m s^{-2}$

So, the deceleration of the car is $7 m s^{-2}$ or we can also say that the acceleration attained by the car to come to rest is $-7 m s^{-2}$.