Question

A truck starts from rest and moves with a constant acceleration of 5.0 m/s2. (a) Find its speed and (b) distance traveled after 4.0 seconds

Answer 1

initial velocity=0m/s
acceleration=5m/s
time=4s
(i)speed acquired,
v=u+at
v=0+5×4
v=20m/s
(ii)distance travelled,
s=ut+1/2at2
s=0×4+1/2×5×4^2
s=5/2×4×4
s=5×8
s=40m.

Answer 2

Explanation:

Initial speed of the truck, u = 0 (at rest)

Acceleration of the truck, $a=5\ m/s^2$

Time, t = 4 s

1. Let v is the final speed of the truck. Using the first equation of motion to find it as :

$v=u+at$

$v=5\times 4$

v = 20 m/s

2. Let s is the distance covered by the truck. It is given by :

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

$s=\dfrac{1}{2}\times 5\times 4^2$

s = 40 meters

Hence, this is the required solution.