Question

a car accelerates uniformlyfrom 18 kilometre per hour to 36 km per hour in 5 seconds calculate (i) the acceleration and(ii) the distance covered by the car in that time ​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Acceleration=1\:m/s^{2}}}}$

$\green{\tt{\therefore{Distance\:covered=37.5\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt:\implies Initial \: velocity(u) = 18 \: km/h \\ \\ \tt:\implies Final\: velocity(v) = 36\: km/h \\ \\ \tt:\implies Time = 5\: sec \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Acceleration(a) = ?\\ \\ \tt: \implies Distance \: travel(s) =?$

• According to given question :

$\tt \circ \: u = 18 \times \frac{5}{18} = 5 \: m/s \\ \\ \tt \circ \: v = 36 \times \frac{5}{18} = 10 \: m/s\\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies v = u + at \\ \\ \tt: \implies 10 = 5 + a \times 5 \\ \\ \tt: \implies 10 - 5 = a \times 5 \\ \\ \tt: \implies a = \frac{5}{5} \\ \\ \green{\tt: \implies a = 1 \: {m/s}^{2} } \\ \\ \bold{As \: we \: know \: that } \\ \tt: \implies s = ut + \frac{1}{2} {at}^{2} \\ \\ \tt: \implies s = 5 \times 5 + \frac{1}{2} \times 1 \times {5}^{2} \\ \\ \tt: \implies s = 25 + 12.5 \\ \\ \green{\tt: \implies s = 37.5 \: m}$

Answer 2

Correct Question :

A car accelerates uniformly from 18 kilometre per hour to 36 km per hour in 5 seconds .

Calculate :

(i) the acceleration .

(ii) the distance covered by the car in that time .

Solution -

In the above Question , the following information is given -

A car accelerates uniformly from 18 kilometre per hour to 36 km per hour in 5 seconds .

So , the innitial Velocity , u of the car -

=> 18 kmph

=> 18 × ( 5 / 18 ) m / s

=> 5 m / s.

Hence ,

U = 5 m/s.

The final Velocity , V of the car -

=> 36 kmph

=> 36 × ( 5 / 18 ) m / s

=> 5 × 2 m / s

=> 10 m / s.

Hence V = 10 m / s.

Time = 5 Seconds

According to the first Equation if motion ,

Acceleration = [ v - u ] / t

=> 1 m / s

According to the second Equation if Motion ,

S = ut + ( 1 / 2 ) a t^2

=> S = 25 + ( 1 / 2 ) × 25

=> S = 75 / 2

=> S = 37.5 m