Question

A particle starts with an initial velocity 2.5 m/s along the positive x-direction and it accelerates uniformly at the rate 0.5 m/s2.

(i) Find the distance travelled by it in the first two seconds.

(ii) how much time does it take to attain the velocity 7.5 m/s ?

(iii) How much distance will it cover in attaining the velocity 7.5 m/s?

Answer 1

u=2.5
acc=0.5
time=2 sec
s=ut+1/2at2
=2.5×2+1/2×0.5×2×2
=5+1
=6metres
final velocity=7.5
a=v-u/t
0.5=7.5-2.5/ t
0.5t=5
t=5/0.5
t=10 sec
s=ut+1/2at2
=2.5×10+1/2×0.5×10×10
=25+25
=50 metres

Answer 2

Answer:

(I). The distance traveled by it in the first two seconds is 6 m.

(II). It takes to attain the velocity 7.5 m/s in 10 sec.

(III). The distance covered in 10 sec is 50 m.

Explanation:

Given that,

Initial velocity = 2.5 m/s

Acceleration = 0.5 m/s²

(I). The distance traveled by it in the first two seconds,

Using second equation of motion,

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

$s=2.5\times2+\dfrac{1}{2}\times0.5\times4$

$s= 6\ m$

(II). We need to calculate the time,

Using equation of motion

$v=u+at$

Put the value in equation

$7.5=2.5+0.5t$

$t=\dfrac{7.5-2.5}{0.5}$

$t=10\ sec$

(III). After 10 sec it attains a speed of 7.5 m/s.

So, the distance covered in 10 sec

Using equation of motion

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

$s=2.5\times10+\dfrac{1}{2}\times0.5\times100$

$s=50\ m$

Hence, (I). The distance traveled by it in the first two seconds is 6 m.

(II). It takes to attain the velocity 7.5 m/s in 10 sec.

(III). The distance covered in 10 sec is 50 m.