Question

A car accelerates uniformly from 54
km/h to 90km/h in 10 seconds . Find the acceleration and the distance traveled by the car in the given time ​

Answer 1

Answer:

Explanation:

INITIAL VELOCITY , U = 54 KM/H

                                                          [ TO CONVERT KM/H TO M/S * BY

                                                                                              5/18]                                                                    

                                 U = 54*5/18

                                     = 15 M/S

FINAL VELOCITY , V = 90 KM/H

                                   = 90*5/18

                                   =  25 M/S.

TIME TAKEN ,T = 10 SEC.

ACCELERATION,A = ?

DISTANCE , S = ?

ACCORDING TO NEWTON'S LAWS OF MOTION:

V = U +AT

25 = 15+A*10

10 = A*10

A = 1 M/S²

S = UT +0.5 A T²

S = 15*10 +0.5*1*100

S = 150+50

S = 200 M

  = 0.2 KM

THEREFORE :

ACCELERATION OF THE BODY = 1 M/S²

DISTANCE TRAVELLED BY THE BODY = 200 M = 0.2 KM.

NOTE :

TO CONVERT M/S TO KM/H MULTIPLY BY 18/5.

HOPE THIS HELPS.

Answer 2

Answer:

a =   $1 m/s^{2}$

s= 200 m

Explanation:

Given,

u = 54 kmph

v = 90 kmph

t = 10 sec

$u = 54 km/h \\ = 54 *\frac{5}{18} = 15 m/s\\ \\v = 90 km/h\\= 90 * \frac{5}{18}\\ 25 m/s$

From the equations of motions

$v = u + at\\and\\s= ut + \frac{1}{2} a t^{2}$

$a = \frac{v-u}{t}\\ =\frac{25- 15}{10}\\ = 1 m/s^{2} \\\\v^{2}= u^{2} + 2as\\625 - 225 = 2* 1*s\\ s = 400 / 2\\= 200 m$