Question

A car moving with constant acceleration travels the distance between two cones placed 58.2 m apart in 2.97 s .

Part A

The cars speed as it passes the second cone is 25 m/s. What was its velocity as it passed the first cone?

Part B

What is the car's acceleration?

Answer 1

Distance travelled, s = 58.2 m
Time taken, t = 2.97 s
Final velocity, v = 25 m/s

Consider a as the acceleration and u as the initial velocity.
Using the formula, s = ut + 1/2at²,
58.2 = 2.97 x u + 0.5 x a x 2.97²
2.97 u + 4.41045 a = 58.2 ...(1)

Using the formula, v = u + at,
25 = u + 2.97 a ...(2)

A)
From (2),
a = (25 - u) / 2.97 ...(3)
Substituting a in (1),
2.97 u + 4.41045 x [25 - u] / 2.97 = 58.2
2.97 u + 37.125 - 1.485 u = 58.2
1.485 u = 21.075
u = 21.075/1.485
= 14.19 m/s

B)
Substituting u in equation (3),
a = (25 - 14.19) / 2.97
= 3.64 m/s²