Question

60.
A car travelling with an initial velocity 15 m s-1 acquires a velocity 25 m s-1 in 5 s
with a constant acceleration. Calculate :
(a) the average velocity :
b() distance in 5s​

Answer 1

Answer:

Explanation:

$u = 15 m/s\\v = 25 m/s\\t = 5 s$

(a)

$V_{avg} = \frac{Total Distance}{Total time}\\\\Acceleration => v = u +at\\ 25 = 15 + a(5)\\\frac{(25 - 15)}{5} = a\\2 m/s^2 = a\\\\Total Distance = ut + 1/2at^2\\15(5) + 1/2(2)(5)^2 = 100m\\\\Total Time = 5s\\\\V_{avg} = \frac{100}{5}\\\\\\V_{avg} = 20 m/s^2$

(b) 100m from the first part