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

a = 1.428 [m/s²]

v₀ = 5 [m/s]

Explanation:

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

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.

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]

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

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer