Question

The car starts from rest and accelerates with an acceleration of 2 m/s^2 for 5 s. It then travels at a constant speed for 20 s, before decelerating at -5 m/s^2 until it reaches zero speed. How much distance did the car cover during this journey.

Answer 1

Answer:

Total distance covered during the journey is 235 m

Solution:

As per the question:

Initial velocity, v = 0 m/s

Acceleration, a = $2 m/s^{2}$

Time, t = 5 s

Now,

For this, we use eqn 2 of motion:

$d = vt + \farc{1}{2}at^{2}$

$d = 0.t + \farc{1}{2}\times 2\times 5^{2} = 25 m$

The final speed of car after t = 5 s is given by:

v' = v + at

v' = 0 + 2(5) = 10 m/s

Now, the car travels at constant speed of 10 m/s for t' = 20 s with a = 0:

$d' = vt + \farc{1}{2}at^{2}$

$d' = 10\times 20 + \farc{1}{2}\times 0\times 20^{2} = 25 m$

d' = 200 m

Now, the car accelerates at a= - 5 $m/s^{2}$ until its final speed, v" = 0 m/s:

$v"^{2} = v'^{2} + 2ad"$

$0 = {10}^{2} + 2\times (- 5)d"$

$100 = 10d"$

d" = 10 m

Total distance covered = d + d' + d"  = 25 + 200 + 10 = 235 m