Q: A sport car accelerate at the rate of 2.80m/s². How long does it take to reach its top speed of 60.0 km/h, starting from rest ?
Answer:
5.95 s.
Explanation:
From Newton's equation of motion,
a = (v-u)/t ................ Equation 1
Where a = acceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time taken to reach the top speed.
Making t the subject of the equation,
t = (v-u)/a.............. Equation 2
Given: a = 2.8 m/s², u = 0 m/s ( from rest), v = 60 km/h = (60×1000)/3600
v = 16.67 m/s.
Substitute into equation 2
t = (16.67-0)/2.8
t = 5.95 s.
Hence the time taken to reach the top speed = 5.95 s.
A sport car accelerate at the rate of 2.80[tex]m/s^{2}[/tex]. How long does it take to reach its top speed of 60.0mil/h, starting from rest ?
9.58 seconds
Using one of the equations of motion;
v = u + at ---------------------(i)
where;
v = final velocity of the car = 60mil/h
u = initial velocity of the car = 0 (since the car is starting from rest)
a = acceleration of the car = 2.80[tex]m/s^{2}[/tex]
t = time taken for the acceleration
(i) First convert the final velocity v = 60mil/h to m/s
1 mile = 1609.34m
60 miles = 60 x 1609.34m = 96560.6m
Also,
1 hour = 3600s
Therefore;
60mil/h = 96569.6m/3600s = 26.82m/s
(ii) substitute the values of v, u and a into equation (i)
=> v = u + at
=> 26.82 = 0 + 2.80t
=> 2.80t = 26.82
=> t = 26.82 / 2.80
=> t = 9.58s
Therefore the time taken to reach its top speed is 9.58s.
