A car starts from rest and travels to the north for 6.00 s with a uniform acceleration of 2.50

m/s2

. The driver then applies the brakes, causing a uniform acceleration of - 2.00 m/s2

. If

the brakes are applied for 4.00 s,

(i) how fast is the car going at the end of the braking period, and [7 m/s] [2 marks]

(ii) how far has the car gone? [89 m] [3 marks]

Question

Physics

Janet Townsend

22 September, 08:04

A car starts from rest and travels to the north for 6.00 s with a uniform acceleration of 2.50

m/s2

. The driver then applies the brakes, causing a uniform acceleration of - 2.00 m/s2

. If

the brakes are applied for 4.00 s,

(i) how fast is the car going at the end of the braking period, and [7 m/s] [2 marks]

(ii) how far has the car gone? [89 m] [3 marks]

+1

Answers (1)

Know the Answer?

Aedan Brewer · Answer 1

To calculate the final velocity, we use Newton's first equation of linear motion:v=u+at

Where v is final velocity

u is initial velocity

a is the average acceleration

t is the time taken during acceleration.

Therefore,

v=0+2.5m/s²*6.00s

=15m/s

Decelerating from 15m/s;

v=15m/s + (-2m/s²*4.0s)

=3m/s To get the distance it travelled, we use v²=u²+2as During acceleration, the distance travelled is calculated as below. 15²=0+2*2.5S 225=5S S=45meters During decellaration, displacement is calculated as below, 3²=15² + (2*4S) 9=225+8S 8S=216 S=27meters Total displacement=45m+27m = 72 meters.