Question

A certain automobile manufacturer claims that its super-deluxe sports car will
accelerate from rest to a speed of 21.0 ms'in 4.0 s. Under the important assumption
that the acceleration is constant:
(a) determine the acceleration of car
(b) find the distance the car travels in 4.0 second
(C) find the distance the car travels in 4 second​

Answer 1

Answer:

(a) Acceleration of the car = 5.25 m/s²

(b) Distance travelled in 4 seconds = 42 m

(c) Distance travelled in 4th second = 18.375 m

Explanation:

velocity changes from zero to 21 m/s in 4 seconds

Therefore, using the first equation of motion

$v=u+at$

Here,

$u=0$

$v=21$ m/s

$t=4$ s

Therefore,

$21=0+a\times 4$

$\implies a=\frac{21}{4}$

$\implies a=5.25$ m/s²

Using the third equation of motion, it distance travelled is s then

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

$\implies 21^2=0^2+2\times \frac{21}{4}\times s$

$\implies s=21\times 2=42$ m

Distance travelled by the car in 3 seconds, using the second equation of motion

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

$s'=0\times 3+\frac{1}{2}\times 5.25\times 3^2$

$\implies s'=23.625$ m

Distance travelled in 4th second

$=42-23.625$

$=18.375$ m

Hope this helps.