Question

1) A car accelerates uniformly
from 18 km h to 36 km h
to 36 km h in 5 s.
Calculate (1) the acceleration and (ii) the
distance covered by the car in that time.​

Answer 1

Answer:

18km/hr = 18x5/18 = 5m/s = u

36km/hr = 36x5/18 = 10m/s = v

t = 5s

a = (v-u)/t

a = 10-5/5

a = 1m/s²

s is the distance covered.

s = ut + 1/2at²

s = (5)(5) + 1/2(1)(5)²

s = 25 + 12.5

s = 37.5m

Hope it helps you!!

Answer 2

☘️ Proper question :

• A car accelerates uniformly from 18 km h^-1 to 36 km h^-1 in 5 s. Calculate (I) the acceleration and (ii) the distance covered by the car in that time.

$\begin{gathered}\\ \\ \\\end{gathered}$

☘️ Required Solution :

We are given that,

• u = 18 km h^-1 = 5 m s^-1
• v = 36 km h^-1 = 10 m s^-1
• and t = 5 s.

$\begin{gathered}\\ \\\end{gathered}$

(i) From Eq. (8.5) we have

$\begin{gathered}\\ \large \sf a = \dfrac{v - u}{t} \\ \\ \sf \: = \frac{10 \: m \: s {}^{ - 1} \: - \: 5 \: m \: s {}^{ - 1} }{5 \: s} \\ \\ \sf \: = 1 \: m \: s {}^{ - 2}\\ \\ \\ \end{gathered}$

(ii) From Eq. (8.6) we have

$\begin{gathered} \\ \sf \: s = u \: t \: + \frac{1}{2} a \: t {}^{2} \\ \\ \sf \: = 5 \: m \: s {}^{ - 1} \times 5 \: s \: + \: \frac{1}{2} \times 1 \: m \: s {}^{ - 2} \: \times \: (5 \: s) {}^{2} \\ \\ \sf \: = 25 \: m \: + \: 12.5 \: m \\ \\ \sf \: = 37.5 \: m \\ \\\end{gathered}$

So, the acceleration of the car is 1 m s^-2 and the distance covered is 37.5 m.