A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of a gazelle assumes an acceleration of 4.2 m/s2m/s2 for 6.5 ss , after which the gazelle continues at a steady speed. What is the gazelles top speed?

v = 27.3 m/s

Explanation:

Given that

Acceleration ,a= 4.2 m/s²

Time ,t= 6.5 s

Lets take the maximum speed gain by Thomson's= v

We know that ,if acceleration is constant then the speed v is given as

v= u +  a t

v=final speed

u=initial speed

a=acceleration

t=time

Here the initial speed of Thomson's ,u = 0 m/s

Now by putting the values in the above equation we get

v= 0 + 4.2 x 6.5 m/s

v = 27.3 m/s

Therefore the maximum speed gain by Thomson will be 27.3 m/s.

54 degrees incident angle = reflected angle

angle is from normal turned to line of ray