EXAMPLE 5-8 Two components of acceleration. A race car starts from rest in the pit area and accelerates at a uniform rate to a speed of 35 m/s in 11 s, moving on a circular track of radius 500 m. Assuming constant tangential acceleration, find (a) the tangential acceleration, and (b) the radial acceler- ation, at the instant when the speed is v = 15 m/s. APPROACH The tangential acceleration relates to the change in speed of the car, and can be calculated as atan = Av/At. The centripetal acceleration relates to the change in the direction of the velocity vector and is calculated using ar = v²/r. SOLUTION (a) During the 11-s time interval, we assume the tangential accel- eration atan is constant. Its magnitude is Δυ (35 m/s – 0m/s) Atan 3.2 m/s². At 11 s (b) When v = 15 m/s, the centripetal acceleration is v² (15 m/s)? ar 0.45 m/s². (500 m) NOTE The radial (centripetal) acceleration increases continually, whereas the tangential acceleration stays constant.
Gravitational force
In nature, every object is attracted by every other object. This phenomenon is called gravity. The force associated with gravity is called gravitational force. The gravitational force is the weakest force that exists in nature. The gravitational force is always attractive.
Acceleration Due to Gravity
In fundamental physics, gravity or gravitational force is the universal attractive force acting between all the matters that exist or exhibit. It is the weakest known force. Therefore no internal changes in an object occurs due to this force. On the other hand, it has control over the trajectories of bodies in the solar system and in the universe due to its vast scope and universal action. The free fall of objects on Earth and the motions of celestial bodies, according to Newton, are both determined by the same force. It was Newton who put forward that the moon is held by a strong attractive force exerted by the Earth which makes it revolve in a straight line. He was sure that this force is similar to the downward force which Earth exerts on all the objects on it.
(I) Determine the tangential and centripetal components
of the net force exerted on the car (by the ground) in
Example 5–8 when its speed is 15 m/s. The car’s mass is 950 kg
![EXAMPLE 5-8 Two components of acceleration. A race car starts from
rest in the pit area and accelerates at a uniform rate to a speed of 35 m/s in 11 s,
moving on a circular track of radius 500 m. Assuming constant tangential
acceleration, find (a) the tangential acceleration, and (b) the radial acceler-
ation, at the instant when the speed is v = 15 m/s.
APPROACH The tangential acceleration relates to the change in speed of the car,
and can be calculated as atan = Av/At. The centripetal acceleration relates to the
change in the direction of the velocity vector and is calculated using ar = v²/r.
SOLUTION (a) During the 11-s time interval, we assume the tangential accel-
eration atan is constant. Its magnitude is
Δυ
(35 m/s – 0m/s)
Atan
3.2 m/s².
At
11 s
(b) When v = 15 m/s, the centripetal acceleration is
v²
(15 m/s)?
ar
0.45 m/s².
(500 m)
NOTE The radial (centripetal) acceleration increases continually, whereas the
tangential acceleration stays constant.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fec639807-6484-485a-a34b-02ba1ca72521%2Fed75a35d-9e88-4dd5-bc79-368b86bc1aa4%2F2psm80q.png&w=3840&q=75)
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