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melamori03 [73]
4 years ago
8

Consider a mass initially moving at 7.50 m/s. How does it take to move 3.5 km (Be sure to convert to meters) if it accelerates a

t .550 m/s^2.?
Please show your work.
Physics
1 answer:
Tomtit [17]4 years ago
5 0
The mass is moving by uniformly accelerated motion, with initial velocity v_i=7.50 m/s and acceleration a=0.55 m/s^2. Its position at time t is given by the following law:
x(t)=v_i t + \frac{1}{2}at^2
where we take the initial position x_i=0 since we are only interested in the distance traveled by the mass.

If we put x(t)=d=3.5 km=3500 m into the equation, the corresponding time t is the time it takes for the mass to travel this distance:
\frac{1}{2}at^2+v_it-d=0
4.9t^2+7.5t-3500 =0
And the two solutions for the equation are:
t=-25.5 s --> negative, we can discard it
t=24.0 s --> this is the solution to our problem
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