Astopwatch starts while race car travels at 4 m/s from the pit area and accelerates at a uniform rate to a speed of 23 m/s in 2 s moving on a circular track of radius 527 m. assuming constant tangential acceleration, find (a) the tangential acceleration, and (b) the radial acceleration, at the instant when the speed is v = 16 once you have both of those find the magnitude of a at any moment

(a) Tangential acceleration:

The tangential acceleration is given by:

where

v = 23 m/s is the final velocity of the car

u = 4 m/s is the initial velocity of the car

t = 2 s is the time taken for the car to accelerate from u to v

Substituting the numbers into the formula, we find

(b) Radial acceleration:

The radial acceleration is equivalent to the centripetal acceleration, which is given by:

where

v = 16 m/s is the tangential speed

r = 527 m is the radius of the circular orbit

Substituting numbers, we find

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1