Question

A car moves along a straight road with constant accelration of 5ms¬². Initially at 5m , its velocity was 3ms¬² compute its position and velocity at t=2s.

Answer 1

Answer:

its position = 21 m and velocity = 13 m/s at t = 2 s

Explanation:

Given:

Acceleration of the car, a  = 5 m/s²

Initial distance = 5 m

initial velocity of the car , u = 3 m/s

Time, t = 2 s

now,

from the Newton's equation of motion, we have

$s=ut+\frac{1}{2}at^2$

where, s is the distance traveled

on substituting the respective values, we get

$s=3\times2+\frac{1}{2}\times5\times2^2$

or

s = 16 m

hence, the final position of the car is = initial distance + Distance covered

or

The final position of the car = 5 + 16 = 21 m

also,

From another Newton's equation of motion

v = u + at

here,

v is the final velocity of the car after t = 2 s

thus, we have

v = 3 + ( 5 × 2 )

or

v = 13 m/s

its position = 21 m and velocity = 13 m/s at t = 2 s