Find the distance traveled during this time and the final velocity of the car

Question

A car has a velocity of $10 ms^{-1}$. It accelerates at $0.2 ms^{-2}$ for half a minute. Find the distance traveled during this time and the final velocity of the car. $(390 m, 16 ms^{-1})$

Accepted Answer

Solution:

Initial velocity = Vi = $10 ms^{-1}$

Acceleration a = 0.2 $ms^{-2}$

Time t = 0.5 min = $0.5 imes60$ = 30 s

(I) Distance S =?

(II) Final velocity Vf =?

S = $Vi imes t + \frac{1}{2} imes a imes t2$

S= $10 imes 30 + \frac{1}{2} imes 0.2 imes (30)^{2}$

S = $300 + \frac{1}{2} imes 0.2 imes 90$

S = $300 + \frac{1}{2} imes \frac{2}{10} imes$ 90

S = 300 + 90

S = 390 m

(II)Using 1st equation of motion

Vf = Vi + at

Vf = $10 + 0.2 imes30$

Vf = 10 + 6

Vf = $16 ms^{-1}$