Question

A car starts from rest and accelerates at 4 m/s^2 for 8 s, after which it continues at constant speed for 19 s. How far would it have traveled from its starting point by that time.

Answer 1

Answer:

The total distance traveled is 736 m

Solution:

According to the question:

Initial velocity, v = 0

(since, the car is starting from rest)

$acceleration, a = 4 m/s^{2}$

Time taken, t = 8 s

Now, the distance covered by it in 8 s is given by the second eqn of motion:

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

$d = 0.t + \farc{1}{2}4\times 8^{2} = 128 m$

Now, to calculate the velocity, we use eqn 1 of motion:

v' = v + at

v' = 0 + 4(8) = 32 m/s

Now, the distance traveled by the car with uniform velocity of 32 m/s for t' = 19 s:

distance, d' = v't'

$d' = 32\times 19 = 608 m$

Total distance traveled = d + d' = 128 + 608 = 736 m