Question

a sport car accelerate at the rate of 2.80m/s. How long does it take to reach its top speed of 60.0ml/h, starting from rest ?

Answer 1

Corrected question:

A sport car accelerate at the rate of 2.80
`m/s^(2)`. How long does it take to reach its top speed of 60.0mil/h, starting from rest ?

Answer:

9.58 seconds

Step-by-step explanation:

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
`m/s^(2)`

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.

Answer 2

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.

Step-by-step 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.