Question

An object is thrown straight up at the speed of 49 metres per second. is the equation to find the speed of the object at every second.

• A. $v=49+9.8t$
  
• B. $v=49-9.8t$
• C. $v=9.8-49t$
• D. $v=9.8+49t$

Answer 1

The correct option is B $v=49-9.8t$

Given that 49 metres per second is the initial speed. That means at t = 0 s, speed must be 49 m/s.
Consider $v=49-9.8t$ and $v=49+9.8t$. In both the cases, at t = 0 s, speed is 49 m/s.

If we throw an object upwards, then speed could be positive, negative and zero. That is why we need a negative sign in the equation.
Example: at t = 5,
$v=49-9.8t$
$v=49-9.8\times 5$
$v=49-49$
$v=0$ m/s

at t = 5,
$v=49+9.8t$
$v=49+49$
$v=98$ m/s
No matter what the value of t, the equation $v=49+9.8t$ will not become 0 or negative. value of v will always increase.
That is why only $v=49-9.8t$ is the correct equation.