Question

A body is thrown vertically upwards with a speed of 20 ms 1. The ratio of magnitude of average velocity and average speed for first 3 seconds of its journey is (g = 10 ms 2)​

Answer 1

Answer:

Solution

verified

Correct option is

A

t=5s

Time to reach maximum height can be obtained from v=u+at

0=20+(−10)t

t=2s

s=ut+0.5at

2

=20(2)+0.5(−10)(2)

2

=20m

Thus, total distance for maximum height is 45 m

s=ut+0.5at

2

45=0+0.5(10)(t

′

)

2

t

′

=3s

Answer 2

Given: A body is thrown vertically upwards with a speed of 20 m/s.

To find: Ratio of magnitude of average velocity and average speed for first three seconds of its journey

Explanation: Initial velocity (u)= 20 m/s

Acceleration due to gravity(g)= 10 m/s^2

Time(t)= 3 second

Let distance covered be s.

$s = ut - \frac{1}{2} g {t}^{2}$

$s = 20 \times 3 - \frac{1}{2} \times 10 \times {3}^{2}$

$s = 60 - 45$

s= 15 m

Distance covered in 3 s= 15 m

Average speed= 15/3

= 5 m/s

Also, displacement is smallest distance between two points which in this case is 15 m

Average velocity= 15/3

= 5 m/s

Since, both velocity and speed are equal their ratio is 1:1.

Therefore, the ratio of magnitude of the average speed and average velocity for the first three seconds of this journey is 1:1