Question

a tiger moving with constant accleration covers the distance between two points 70 meter apart in 7 seconds . its speed as it pass the second point 15 meter per second . 1) what is the speed at the first point. 2) what us its accleration​

Answer 1

Answer:

a = 1.428 [m/s²]

v₀ = 5 [m/s]

Explanation:

To solve this problem we must use the following equation of kinematics.

$x=x_{o}+v_{o}*t+\frac{1}{2}*a*t^{2}$

where:

x = final point [m]

x₀ = initial point [m]

v₀ = initial velocity [m/s]

a = acceleration [m/s²]

t = time [s]

But we need to use this additional equation.

$v_{f}=v_{o}+a*t$

where:

vf = final velocity = 15 [m/s]

Now we can use this equation, replacing it, in the first one. We must bear in mind that the difference among x - x₀ is equal to 70 [m]

$x-x_{o}=v_{o}*t+\frac{1}{2}*a*t^{2} \\x-x_{o}=(v_{f}-a*t)*t+\frac{1}{2} *a*t^{2}\\70=(15-a*t)*t+\frac{1}{2}*a*t^{2}\\70=15*t-a*t^{2} +\frac{1}{2}*a*t^{2} \\70=15*t-\frac{1}{2}*a*t^{2}\\70=15*(7)-\frac{1}{2} *a*(7)^{2}\\105-70=0.5*a*49\\35=24.5*a\\a=1.428[m/s^{2} ]$

Now replacing this value in the second equation, we can find the initial velocity.

$15=v_{o}+1.428*7\\v_{o}=5[m/s]$