Question

A truck from rest ana accelerate at3m s² for 5sec and then continue with constant velocity calculate the distance covered in 15 seconds since it start from rest ​

Answer 1

so, the truck start from rest and then with a constant acceleration of 3ms$^2$ for 5 sec and the velocity after the third second was constant and we have to calculate the total distance covered by 15 sec of start. By 5 sec is the time the truck accelerated.

now, u = 0

v = ? (to be find)

t = 5 sec

a = 3 ms$^{-2}$

$a = \frac{v - u}{t}$

$3ms^{-2} = \frac{v - 0}{5}$

$15ms^{-1} = v$

so, now we know the speed after which the truck move with the constant speed.

now, to find the distance

S = ? (to be find)

U = 0

T = 5

A = 3ms$^{-2}$

now the equation

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

$S = 0 + \frac{1}{2} \times 3 \times 5^2$

$S = 37.5$

so, this time the truck travelled 37.5 meters and then he moved with a constant speed which was the speed after the acceleration.

so, now the distance with the constant speed of 15 m/s

$Distance = Speed \times Time$

$Distance = 15 \times 10$

$Distance = 150$

We take the time 10 because the truck was accelerating five seconds and we have to find the total distance in 15 second also that the speed was constant after 5 second.