Question

1. A car accelerates from 6ms -1 to 16 ms -1 in
10 seconds. Calculate
a)the acceleration and
(b)the distance covered by the car in that time.

Answer 1

Answer:

acceleration = (v-u)/t = (16-6)/10= 1 m/s sq

Distance= u + 1/2 a * t sq

= 6 + 1/2* 1 * 10 sq

= 6 + 1/2*100

= 6 + 50

= 56 m

Answer 2

Given:-

initial velocity = 6 m/s

final velocity = 16 m/s

time = 10 seconds

To find:-

a) acceleration

b) distance

Procedure:-

$acceleration = \frac{final \: velocity - initial \: velocity}{time}$

acceleration = ( 16 - 6 )/10

acceleration = 10/10 = 1 m/s²

distance = initial velocity×time + 1/2acceleration×time²

distance = ( 6×10 ) + ( 1/2 × 1 × 10² )

distance = 60 + 50

distance = 110 m